Equivalence of Two Norm and Infinity Norm

For the RHS (since, for all $i$, $|x_i|\leq \sup_j |x_j|:=\|x\|_{\infty} \ \Rightarrow \ x_i^2 \leq \|x\|_{\infty}^2$) : \begin{align*} \|x\|_2=\sqrt{\sum_i x_i^2}& \leq \sqrt{\sum_i \|x\|_{\infty}^2} \\ & =\sqrt{n \|x\|_{\infty}^2}=\sqrt{n} \|x\|_{\infty} \end{align*}

Not really an answer, but too long for a comment

The proof of your LHS is not correct. Given a vector $x = (x_1, \ldots, x_n)$, we have the following definitions: $$\|x\|_\infty = \max \{|x_i|\}$$ and $$\|x\|_2 = \sqrt{\sum x_i^2} = \sqrt{\sum |x_i|^2}.$$ Hence we find that $$\|x\|_\infty^2 \leq \sum |x_i|^2 = \|x\|_2^2.$$ The RHS is proven in a similar way, as has been shown in other answers.