Yes, (b) implies (a) also in the case of non-separable inner product spaces. Assuming $X$ is not complete we construct below a maximal orthonormal set which is not dense in $X$. You need the Axiom of Choice to get the maximal set but with a little twist added in order to show maximality. Two useful lemmas:

Lemma 1: If $(u_n)_n$ is a Cauchy sequence in $X$ then $\ell(x)=\lim_n \langle u_n, x\rangle$ defines a continuous linear functional of $x\in X$. If $\ker\ell = X$ then $u_n \to 0$ (i.e. a Cauchy sequence that goes weakly to zero, converges to zero). Proof-hint: $(\|u_n\|)_n$ is a Cauchy sequence in ${\Bbb R}_+$.

Lemme 2: Let $(u_n)_n$ be a Cauchy sequence with $\|u_n\|=1$ but which does not converge in $X$ and let $\ell$ be the associated linear functional. Then $M=\ker\ell$ is a proper closed subspace of $X$ and $M^\perp = \{0\}$. Proof-hint: If $M^\perp$ contains a non-trivial element $z$ use Lemma 1 to show that $u_n\to c z$ for some constant $c$, contradicting the hypotheses.

Now, consider the collection $\Lambda$ of orthonormal subsets of $M$ with a partial order given by inclusion. Then $\Lambda$ is inductively ordered and by Zorn, $\Lambda$ has a maximal element which we denote ${\cal B}$. Let $N\subset M$ be the closure of the span of ${\cal B}$. Several cases must be considered.

1. If $N=M$ the proof is over by Lemma 2. ${\cal B}$ is maximal but does not generate $X$.

2. if $N^\perp=\{0\}$ the proof is also over (same reasons).

3. If $N^\perp\cap M\neq \{0\}$ that would contradict maximality of ${\cal B}$ so we may assume that this intersection is trivial.

4. If $N^\perp$ contains at least two independent elements $p_1,p_2$ then $q=p_1/\ell(p_1)-p_2/\ell(p_2) \in N^\perp\cap M$, again contradicting maximality of ${\cal B}$.

5. There remains only one case to consider, i.e. that $N^\perp$ is one-dimensional, whence generated by precisely one non-trivial element $p\in X\setminus M$ and that $M\setminus N$ contains some non-zero vector $q$. Let ${\cal B}'={\cal B} \cup \{p\}$ and let $H$ be the closed span of ${\cal B}'$. By orthonormality, it is the direct sum of $N$ and the span of $p$. First, we must have $H^\perp=\{0\}$ so ${\cal B}'$ is a maximal orthonormal set in $X$. Second, suppose that $q\in H$. Then $q=cp + n$ with $c$ a constant and $n\in N$. But as $\ell(p)\neq 0$ and $0=\ell(q)=c \ell(p) + \ell(n) = c \ell(p)$ we wee that $c=0$ so $q\in N$ contrary to hypothesis. Thus, the orthonormal set ${\cal B}'$ must be maximal but not generating.

Yes, (b) implies (a) also in the case of non-separable inner product spaces. Assuming $X$ is not complete we construct below a maximal orthonormal set, the span of which is not dense in $X$. You need the Axiom of Choice to get the maximal set but with a little twist added in order to show maximality. Two useful lemmas:

Lemma 1: If $(u_n)_n$ is a Cauchy sequence in $X$ then $\ell(x)=\lim_n \langle u_n, x\rangle$ defines a continuous linear functional of $x\in X$. If $\ker\ell = X$ then $u_n \to 0$ (i.e. a Cauchy sequence that goes weakly to zero, converges to zero). Proof-hint: $(\|u_n\|)_n$ is a Cauchy sequence in ${\Bbb R}_+$.

Lemme 2: Let $(u_n)_n$ be a Cauchy sequence with $\|u_n\|=1$ but which does not converge in $X$ and let $\ell$ be the associated linear functional. Then $M=\ker\ell$ is a proper closed subspace of $X$ and $M^\perp = \{0\}$. Proof-hint: If $M^\perp$ contains a non-trivial element $z$ use Lemma 1 to show that $u_n\to c z$ for some constant $c$, contradicting the hypotheses.

Now, consider the collection $\Lambda$ of orthonormal subsets of $M$ with a partial order given by inclusion. Then $\Lambda$ is inductively ordered and by Zorn, $\Lambda$ has a maximal element which we denote ${\cal B}$. Let $N\subset M$ be the closure of the span of ${\cal B}$. Several cases must be considered.

1. If $N=M$ the proof is over by Lemma 2. ${\cal B}$ is maximal but does not generate $X$.

2. if $N^\perp=\{0\}$ the proof is also over (for the same reason).

3. If $N^\perp\cap M\neq \{0\}$ that would contradict maximality of ${\cal B}$ so we may assume that this intersection is trivial.

4. If $N^\perp$ contains at least two independent elements $p_1,p_2$ then $q=p_1/\ell(p_1)-p_2/\ell(p_2) \in N^\perp\cap M$, again contradicting maximality of ${\cal B}$.

5. There remains just one case to consider, i.e. that $N^\perp$ is one-dimensional, whence generated by one unit vector $p\in X\setminus M$ and that $M\setminus N$ contains some non-zero vector $q$. Let ${\cal B}'={\cal B} \cup \{p\}$ and let $H$ be the closed span of ${\cal B}'$. By orthonormality, it is the direct sum of $N$ and the span of $p$. First, we must have $H^\perp=\{0\}$ so ${\cal B}'$ is indeed a maximal orthonormal set in $X$. Second, suppose that $q\in H$. Then $q=cp + n$ with $c$ a constant and $n\in N$. But as $\ell(p)\neq 0$ and $0=\ell(q)=c \ell(p) + \ell(n) = c \ell(p)$ we wee that $c=0$ so $q\in N$ contrary to hypothesis. Thus, the orthonormal set ${\cal B}'$ must be maximal but not generating.

Yes, it also works in non-separable inner product spaces. You need the Axiom of Choice to get a maximal basis but with a little twist to show maximality.

Lemma 1: If $(u_n)_n$ is a Cauchy sequence in $V$ then $\ell(x)=\lim_n \langle u_n, x\rangle$ defines a continuous linear functional of $x\in V$. If $\ker\ell = V$ then $u_n \to 0$ (i.e. a Cauchy sequence that goes weakly to zero, converges to zero). Proof-hint: $(\|u_n\|)_n$ is a Cauchy sequence in ${\Bbb R}_+$.

Lemme 2: Let $(u_n)_n$ be a Cauchy sequence with $\|u_n\|=1$ but which does not converge in $V$ and let $\ell$ be the associated linear functional. Then $M=\ker\ell$ is a proper closed subspace of $V$ and $M^\perp = \{0\}$. Proof-hint: If $M^\perp$ contains a non-trivial element $z$ use Lemma 1 to show that $u_n\to c z$ for some constant $c$, contradicting the hypotheses.

Now, consider the collection $\Lambda$ of orthonormal subsets of $M$ with a partial order given by inclusion. Then $\Lambda$ is inductively ordered and by Zorn, $\Lambda$ has a maximal element which we denote ${\cal B}$. Let $N\subset M$ be the closure of the span of ${\cal B}$. Several cases must be considered.

1. If $N=M$ the proof is over by Lemma 2. $N$ is maximal but does not generate $V$.

2. if $N^\perp=\{0\}$ the proof is also over (same reasons).

3. If $N^\perp\cap M\neq \{0\}$ that would contradict maximality of ${\cal B}$ so we may assume that this intersection is empty.

The above shows the difference compared to a usual proof of existence of basis of $M$, namely that $N^\perp$ need not have non-trivial elements in $M$ even though it may be a proper subspace.

4. If $N^\perp$ contains at least two independent elements $p_1,p_2$ then $q=p_1/\ell(p_1)-p_2/\ell(p_2) \in N^\perp\cap M$, again contradicting maximality.

5. The remaining case to consider is that $N^\perp$ is generated by precisely one non-trivial element $p\in V\setminus M$ and that $M\setminus N$ contains some vector $q$. Let ${\cal B}'={\cal B} \cup \{p\}$ and let $H$ be the closed span of ${\cal B}'$. By orthonormality, it is the direct sum of $N$ and the span of $p$. First, $H^\perp=\{0\}$ so it is a maximal orthonormal set in $V$. Second, suppose that $q\in H$. Then $q=cp + n$ with $c$ a constant and $n\in N$. But as $\ell(p)\neq 0$ and $0=\ell(q)=c \ell(p) + \ell(n) = c \ell(p)$ we wee that $c=0$ so $q\in N$ contrary to hypothesis. Thus, $H$ is maximal but not generating.

Yes, (b) implies (a) also in the case of non-separable inner product spaces. Assuming $X$ is not complete we construct below a maximal orthonormal set which is not dense in $X$. You need the Axiom of Choice to get the maximal set but with a little twist added in order to show maximality. Two useful lemmas:

Lemma 1: If $(u_n)_n$ is a Cauchy sequence in $X$ then $\ell(x)=\lim_n \langle u_n, x\rangle$ defines a continuous linear functional of $x\in X$. If $\ker\ell = X$ then $u_n \to 0$ (i.e. a Cauchy sequence that goes weakly to zero, converges to zero). Proof-hint: $(\|u_n\|)_n$ is a Cauchy sequence in ${\Bbb R}_+$.

Lemme 2: Let $(u_n)_n$ be a Cauchy sequence with $\|u_n\|=1$ but which does not converge in $X$ and let $\ell$ be the associated linear functional. Then $M=\ker\ell$ is a proper closed subspace of $X$ and $M^\perp = \{0\}$. Proof-hint: If $M^\perp$ contains a non-trivial element $z$ use Lemma 1 to show that $u_n\to c z$ for some constant $c$, contradicting the hypotheses.

Now, consider the collection $\Lambda$ of orthonormal subsets of $M$ with a partial order given by inclusion. Then $\Lambda$ is inductively ordered and by Zorn, $\Lambda$ has a maximal element which we denote ${\cal B}$. Let $N\subset M$ be the closure of the span of ${\cal B}$. Several cases must be considered.

1. If $N=M$ the proof is over by Lemma 2. ${\cal B}$ is maximal but does not generate $X$.

2. if $N^\perp=\{0\}$ the proof is also over (same reasons).

3. If $N^\perp\cap M\neq \{0\}$ that would contradict maximality of ${\cal B}$ so we may assume that this intersection is trivial.

4. If $N^\perp$ contains at least two independent elements $p_1,p_2$ then $q=p_1/\ell(p_1)-p_2/\ell(p_2) \in N^\perp\cap M$, again contradicting maximality of ${\cal B}$.

5. There remains only one case to consider, i.e. that $N^\perp$ is one-dimensional, whence generated by precisely one non-trivial element $p\in X\setminus M$ and that $M\setminus N$ contains some non-zero vector $q$. Let ${\cal B}'={\cal B} \cup \{p\}$ and let $H$ be the closed span of ${\cal B}'$. By orthonormality, it is the direct sum of $N$ and the span of $p$. First, we must have $H^\perp=\{0\}$ so ${\cal B}'$ is a maximal orthonormal set in $X$. Second, suppose that $q\in H$. Then $q=cp + n$ with $c$ a constant and $n\in N$. But as $\ell(p)\neq 0$ and $0=\ell(q)=c \ell(p) + \ell(n) = c \ell(p)$ we wee that $c=0$ so $q\in N$ contrary to hypothesis. Thus, the orthonormal set ${\cal B}'$ must be maximal but not generating.