I am able to prove that one curve in $C^1([0,1], \mathbb{R})$, $r$, that has minimal length connecting two points is a straight line. If these points are $p, q \in \mathbb{R}^n$, then the distance is $|q-p|$. I am, however, struggling to prove that no curves other than a straight line have length $|q-p|$ and have no idea where to start.
For a parameterised curve, $r \in C^1([0,1], \mathbb{R})$, I am finding the length via $\ell(r) = \int_0^1 |\dot{r}(t)|dt$.
If $s\in [0,1]$ such that $r(s)$ does not lie on the line segment connecting $r(0)$ and $r(1)$, then by the triangle inequality and the fundamental theorem of calculus we have
\begin{align*} \ell(r) &= \int_0^1 |\dot{r}(t)|dt= \int_0^s |\dot{r}(t)|dt + \int_s^1 |\dot{r}(t)|dt\\ &\geq \left|\int_0^s \dot{r}(t)dt\right| + \left|\int_s^1 \dot{r}(t)dt\right| \\ &= |r(s)-r(0)|+|r(1)-r(s)|>|r(1)-r(0)|. \end{align*}
Remark
Another way to see this is that once we have a definition of length that always satisfies $\ell(r)\geq |r(b)-r(a)|$ for a curve parametrized by $[a,b]$ (which you have already shown apparently, presumably using the fundamental theorem), and satisfies $$\ell(r|_{[a,c]})+\ell(r|_{[c,b]})=\ell(r)$$ for $a<c<b$, then we are done, since if $r(c)$ is not on a line segment with $r(a)$ and $r(b)$ then we have
\begin{align*} \ell(r) &= \ell(r|_{[a,c]})+\ell(r|_{[c,b]})\\ &\geq |r(c)-r(a)|+ |r(b)-r(c)|\\ &>|r(b)-r(a)|. \end{align*}
This is useful to keep in mind when considering curves not parametrized smoothly, using more general definitions of arc-length.