This answer will only deal with the forward implication since you have a good answer dealing with the other one.

First, let's show that weak convergence implies boundedness of norms. This is in fact true in Banach spaces in general by essentially the same idea but I will work in this special case.

Recall that $L^p(X)^*$ is isometrically isomorphic to $L^q(X)$ where $p^{-1} + q^{-1} = 1$. In particular, we can identify each $f_n$ with a linear functional $\phi_n$ on $L^q$ defined by $$\phi_n(g) = \int_X f_n g d\mu$$ and have $\|f_n\| = \|\phi_n\|$.

Now we seek to apply the uniform boundedness theorem to the family $\{\phi_n\}_{n \geq 1}$. To do this, notice that for fixed $g \in L^q(X)$, $\phi_n(g) \to \int_X fg d\mu$ by weak convergence of $f_n$. Since convergent sequences in $\mathbb{R}$ (or $\mathbb{C}$) are bounded this implies that $|\phi_n(g)|$ is a bounded sequence for each $g \in L^q(X)$.

In turn, by the Uniform Boundedness Theorem, we have that $\sup_n \|f_n\| = \sup_n \|\phi_n\| < \infty$.

The second part is immediate since $1_A \in L^q(X)$ so that $$\int_A f_n d \mu = \int_X 1_A f_n d \mu \to \int_x 1_A f d \mu = \int_A f d \mu$$ by the weak convergence.

This answer will only deal with the forward implication since you have a good answer dealing with the other one.

First, let's show that weak convergence implies boundedness of norms. This is in fact true in Banach spaces in general by essentially the same idea but I will work in this special case.

Recall that $L^p(X)^*$ is isometrically isomorphic to $L^q(X)$ where $p^{-1} + q^{-1} = 1$. In particular, we can identify each $f_n$ with a linear functional $\phi_n$ on $L^q$ defined by $$\phi_n(g) = \int_X f_n g d\mu$$ and have $\|f_n\| = \|\phi_n\|$.

Now we seek to apply the uniform boundedness theorem to the family $\{\phi_n\}_{n \geq 1}$. To do this, notice that for fixed $g \in L^q(X)$, $\phi_n(g) \to \int_X fg d\mu$ by weak convergence of $f_n$. Since convergent sequences in $\mathbb{R}$ (or $\mathbb{C}$) are bounded this implies that $|\phi_n(g)|$ is a bounded sequence for each $g \in L^q(X)$.

In turn, by the Uniform Boundedness Theorem, we have that $\sup_n \|f_n\| = \sup_n \|\phi_n\| < \infty$.

The second part is immediate since $1_A \in L^q(X)$ so that $$\int_A f_n d \mu = \int_X 1_A f_n d \mu \to \int_X 1_A f d \mu = \int_A f d \mu$$ by the weak convergence.