Please can you criticize my proof. Not sure if my assumptions are correct.
Show that the metric space $(\mathbb{R}^{n \times m}, \| \cdot \|_\infty)$ is complete.
Let $(a_{p})_{p \in \mathbb{N}}$ be a Cauchy sequence in $\mathbb{R}^{n \times m}$.
Then $\exists N \in \mathbb{N}$ such that $\forall p, q > N, \|a_{p} - a_{q}\|_\infty < 1 $
Choose $q = N+1$. Then $\|a_{p} - a_{N+1}\|_\infty < 1$
$\|a_p\|_\infty = \|a_p\ - a_{N+1} + a_{N+1}\|_\infty \leq \|a_p\ - a_{N+1}\|_\infty + \|a_{N+1}\|_\infty \leq 1 + \|a_{N+1}\|_\infty $
Take $M = max\{{\|a_1\|_\infty, ..., \|a_N\|_\infty, 1 + \|a_{N+1}\|_\infty}\}$. Then $0 \leq \|a_p\|_\infty \leq M$, so $\|a_p\|_\infty$ is bounded.
By the Bolzano-Weierstrass Theorem, there exists a convergent subsequence $(a_{p_k})_{k \in \mathbb{N}}$ in $\mathbb{R}^{n \times m}$.
As $(a_{p})_{p \in \mathbb{N}}$ is Cauchy, $(a_{p})_{p \in \mathbb{N}}$ is also convergent.
If you want you can also use the equivalence of norms.
We know that exist $M,N>0$ such that $$M||x||_2 \leq ||x||_{\infty}\leq N||x||_2$$ $\forall x \in \mathbb{R}^d$
where $||.||_2$ is the Euclideian norm.
In general using the fact that in a finite dimensional space,all norms are equivalent,you can prove completness of $\mathbb{R}^d$ with respect to a given norm $||.||_a$ by proving the completness with respect to a $\text{''simpler''}$ norm.