Let $S\subseteq\mathbb{R}^n$ be a non-empty set. Consider that the set $C_S=\{\gamma:[0,1]\rightarrow\mathbb{R}^n|\gamma\text{ is continuous }, S\subseteq\gamma([0,1]), \gamma \text{ is Lipschitz}\}$ is non-empty and define:
$$\|\gamma\|_{\text{Lip}}=\sup_{s\neq t} \frac{\|\gamma(s)-\gamma(t)\|}{|s-t|}$$
for each $\gamma\in C_S$. I want to show that there exists $\gamma_\star\in C_S$ such that $\|\gamma_\star\|_{\text{Lip}}=\inf \{\|\gamma\|_{\text{Lip}} | \gamma\in C_S\}$.
My first attempt was to try to show, using Arzelà-Ascoli and such, that the set $C_S$ is compact in $C([0,1],\mathbb{R}^n)$ with the $\|f\|_{\infty}$ metric, and then try to construct a continuous function $F:C_S\rightarrow\mathbb{R}$ (with the obvious candidate being $F(\gamma)=\|\gamma\|_{\text{Lip}}$, although I don't know if it is continuous) such that I can guarantee the infimum property.
However, the problem I found is that I cannot prove that $\{\|\gamma\|_{\text{Lip}} | \gamma\in C_S\}$ is bounded, hence there can be arbitrary big Lipschitz constants, what makes it difficult to prove that $C_S(t)=\{\gamma(t)|\gamma\in C_S \}$ is bounded and also that $C_S$ is closed.
I ran out of ideas for the moment, can anyone help?
It doesn't matter if your set of Lipschitz norms is unbounded above, since you're looking for an infimum, and we clearly have the bound from below given by 0.
So take a minimizing sequence, i.e. a sequence of curves $\gamma_k$ such that $$ \| \gamma_k \|_{lip} \downarrow \inf\{ \| \gamma\|_{lip}: \gamma\in C_S\}. $$ You can always find this sequence by definition of infimum. They will have uniformly bounded norms, and your Arzela-Ascoli argument will give the result.