If $\gamma ' (t_0) \ne 0$ then, by the continuity of $\gamma'$, there is a whole neighbourhood $U$ of $t_0$ such that $\gamma ' (t) \ne 0$ on $U$.
Choose then at each point $\gamma (t)$ with $t \in U$ a vector $n(t)$ (a unit normal) such that the couple $\{\gamma ' (t), n(t) \}$ should be a positively oriented orthonormal basis. Let $A(t)$ be the orthogonal matrix making the transition from $\left\{ \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\}$ to $\{ \gamma ' (t), n(t) \}$.
Consider the map $F = (F_x, F_y) : U \times \Bbb R \to \Bbb R^2$ given by $(t, s) \mapsto \gamma (t) + s n(t)$. In words: starting at the point $\gamma(t)$ move along the normal for a length of $s$. The curves $s \mapsto F(t,s)$ allow you to walk along the normal line at $\gamma (t)$, while the curves $t \mapsto F(t,s)$ allow you to walk on a curve homothetic to $\gamma$, and parallel to it at distance $s$ from $\gamma$ measured along the normal at each point, something like "level curves" in the plane, if you want.
Notice that $F(t, 0) = \gamma (t)$, which proves (i).
For the Jacobian we have that
$$\begin{pmatrix} \frac {\partial F_x} {\partial t} & \frac {\partial F_x} {\partial s} \\ \frac {\partial F_y} {\partial t} & \frac {\partial F_y} {\partial s} \end{pmatrix} = \begin{pmatrix} \gamma_x ' (t) + s n_x ' (t) & n_x (t) \\ \gamma_y ' (t) + s n_y ' (t) & n_y (t) \end{pmatrix} ,$$
so evaluating at $(t, 0)$ gives
$$\begin{pmatrix} \frac {\partial F_x} {\partial t} & \frac {\partial F_x} {\partial s} \\ \frac {\partial F_y} {\partial t} & \frac {\partial F_y} {\partial s} \end{pmatrix} (t, 0) = \begin{pmatrix} \gamma_x ' (t) & n_x (t) \\ \gamma_y ' (t) & n_y (t) \end{pmatrix} = A(t) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A(t) ^{-1}$$
which, when letting $t=t_0$, shows that $\Bbb d _{(t_0, 0)} F$ is an invertible linear map, therefore by the inverse function theorem $F$ must be a local diffeomorphism between some neighbourhood $V$ of $(t_0, 0)$ contained in $U \times \Bbb R$ and some neighbourhood of $\gamma (t_0)$.
Since $\det \ \Bbb d _{(t, 0)} F = \det \ (A(t) I_2 A(t)^{-1}) = 1 >0$ for all $(t,0) \in V$, by continuity (and the fact that on $V$ the Jacobian of $F$ is never $0$, because $F \big| _V$ is a diffeomorphism) it follows that $\det \ \Bbb d _{(t, s)} F > 0$ for all $(t,s) \in V$, which proves (ii).
The implicit function theorem does not show up explicitly here, but since it is equivalent to the inverse function theorem (that we have made essential use in the proof), there surely exist a proof based upon it. In fact, every proof based upon any of these two theorems can be converted into a proof based upon the other one, but I shall not attempt to do it here simply because I consider my proof based upon the inverse function theorem to exhibit the geometrical underpinnings in a clearer way.
