Distance of two points in $\mathbb{R}^N$ $\leq$ length of curve

This can be easily seen with a straightforward calculation:

$$\| y - x \| = \| \gamma (1) - \gamma (0)\| = \left \| \int_0^1 \dot{\gamma}(t)\ \mathrm dt \right \| \leq \int_0^1 \|\dot{\gamma}(t)\| \ \mathrm d t = \ell(\gamma). $$

The length of a $C^0$ curve $\gamma\colon[0,1]\to\Bbb R^n$ is defined as the supremum over all $$\|\gamma(t_1)-\gamma(t_0)\|+\|\gamma(t_2)-\gamma(t_1)\|+\ldots+\|\gamma(t_m)-\gamma(t_{m-1})\| $$ where the supremum is taken over all $m$ and ordered sequences $0=t_0<t_1<\ldots <t_m=1$ (and the curve is called rectifiable if that supremum is finite). As $m=1$, $t_0=0$, $t_1=1$ is among these, the claim follows.

You can approximate the curve with a polygonal path (I will define what I mean by "polygonal path" here); by choosing $n$ points $\{ y_k \}_{k=1}^{n}$ on the curve, including its initial and terminal points, approximate it by the polygonal path made of the line segments with endpoints $y_k$ and $y_{k+1}$, $k=1,...,n-1$. As $n\to\infty$, the approximation is better and better. This is because $\gamma$ is $C^1$.

Then we only need to prove that of all the polygonal paths from $x$ to $y$, the shortest one is the one made of only a single line segment, the line segment from $x$ to $y$

Can you take it from there, using the Triangle Inequality and some induction?