Let $S\subset \mathbb{R}^n$ be closed w.r.t. $\lVert\cdot\rVert_\infty$. Then $S$ contains all its limit points under $\lVert\cdot\rVert_\infty$, which means that if $\{x_n\}\subset S$ is a sequence converging to some point $x$, then $x\in S$. More precisely, $\forall\varepsilon>0$, $\exists N>0$ such that $n>N$ implies $\lVert x_n - x \rVert_\infty<\varepsilon$.
Consider now a limit point $x$ of $S$, under $\lVert\cdot\rVert_\infty$. Then $x\in S$, with $\{x_n\}$ being a sequence entirely contained in $S$ and converging to $x$ under $\lVert\cdot\rVert_\infty$. Now, $\lVert x_n - x \rVert_2\le \lVert x_n - x \rVert_\infty=\sup\{\lVert x_n - x \rVert_2\}<\varepsilon$.
But now that's where I'm stuck. I think the next step should be something like this: suppose $x\not\in S$ under $\lVert \cdot \rVert_2$. What is required to be shown in order to arrive at a contradiction? That is, to show that $x$ must be in $S$ under $\lVert \cdot \rVert_2$.
I would go about it like this:
$S$ is closed under $\lVert\cdot\rVert_\infty$ implies its complement $S^c$ is open.
For arbitrary $x\in S^c$ we can find an $\epsilon>0$ so that $S$ contains the open ball under $\lVert\cdot\rVert_\infty$, which is $B_\infty(x,\epsilon)\subseteq S^c$
However, $\lVert x-y\rVert_\infty<\epsilon$ implies $\frac{1}{\sqrt n}\lVert x-y\rVert_2\le\lVert x-y\rVert_\infty<\epsilon$
So $B_2(x,\epsilon\sqrt n)\subseteq B_\infty(x,\epsilon)\subseteq S^c$
So $S^c$ is open under Euclidean norm, so $S$ is closed under Euclidean norm.