Suppose $\gamma \in C^1(\mathbb{R} , \mathbb{R}^2)$ satisfies $$ ||\gamma|| = 1$$ and $$ ||\gamma ' || = 1$$ where $|| \cdot ||$ is the usual Euclidean norm on $\mathbb{R}^2$, and $\gamma'$ is the derivative of $\gamma$.
Is there a way to show that $\gamma(\mathbb{R}) = S^1$ and that $\gamma$ is periodic directly from the above equations?
I think the following works:
Your conditions imply the slightly more convenient conditions $||\gamma(t)||^2 = 1$ and $||\gamma'(t)||^2 = 1$ for all $t$.
$||\gamma(t)||^2 = \gamma(t) \cdot \gamma(t) = 1$
By the product rule, $2 \gamma'(t) \cdot \gamma(t) = 0 \implies \gamma'(t) \cdot \gamma(t) = 0 \implies x(t)x'(t)+y(t)y'(t) = 0$
Explicitly the system of equations is $\begin{cases} x(t)x'(t)+y(t)y'(t) = 0 \\ x'(t) x'(t) + y'(t)y'(t) = 1 \end{cases}$
$\begin{pmatrix} x(t) & y(t) \\ x'(t) & y'(t) \end{pmatrix} \begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$
So $\begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix} = \frac{1}{x(t)y'(t) - y(t)x'(t)} \begin{pmatrix} y'(t) & -y(t) \\ -x'(t) & x(t) \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ i.e. $\begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix} = \frac{1}{x(t)y'(t) - y(t)x'(t)} \begin{pmatrix} -y(t) \\ x(t) \end{pmatrix}$
Note that by the identity $( \gamma \times \gamma' )^2 = ||\gamma||^2||\gamma'||^2 - (\gamma \cdot \gamma')^2$ (which is readily checked --- here the cross product is the two-dimensional version), it follows that $\gamma \times \gamma' = x(t)y'(t) - y(t)x'(t) = \pm 1$
The system of equations $\begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix} = \begin{pmatrix} -y(t) \\ x(t) \end{pmatrix}$ has the family of solutions $x(t) = \cos(t+\phi), y(t)=\sin(t+\phi)$ (counterclockwise).
Similarly, the system of equations $\begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix} = - \begin{pmatrix} -y(t) \\ x(t) \end{pmatrix}$ has the family of solutions $x(t) = \cos(-t+\phi), y(t)=\sin(-t+\phi)$ (clockwise).
So it seems the conditions are enough to imply $2\pi$-periodicity even.