$\newcommand\R{\mathbb R}$

1. It does not make sense to define a convex function on the nonconvex set $[-\infty,\infty]^m$.

2. Even if you replace $[-\infty,\infty]^m$ by $\mathbb R^m$ and even you just take $m=1$, then your desired statement will still be false in general. For instance, do take $m=1$ and let $f(x):=e^x$ and $f_n(x):=e^x\,1(x\ge-n)+e^{-n}(x+n+1)\,1(x<-n)$ for real $x$ and natural $n$. Then $f$ and the $f_n$'s are convex, $f_n\to f$ pointwise on $\R$ (as $n\to\infty$), but $$\inf_\R f_n=-\infty\not\to0=\inf_\R f.$$

A necessary and sufficient condition for the infimum-stability for convex functions $f$ defined on $\R$ was given by Theorem 1 (one can also use the arXiv version).

The proof of the just mentioned result takes about 4 pages. No extension of that result to $m>1$ seems to be known.

1. $\newcommand\R{\mathbb R}$It does not make sense to define a convex function on the nonconvex set $[-\infty,\infty]^m$.

2. Even if you replace $[-\infty,\infty]^m$ by $\mathbb R^m$ and even you just take $m=1$, then your desired statement will still be false in general. For instance, do take $m=1$ and let $f(x):=e^x$ and $f_n(x):=e^x\,1(x\ge-n)+e^{-n}(x+n+1)\,1(x<-n)$ for real $x$ and natural $n$. Then $f$ and the $f_n$'s are convex, $f_n\to f$ pointwise on $\R$ (as $n\to\infty$), but $$\inf_\R f_n=-\infty\not\to0=\inf_\R f.$$

A necessary and sufficient condition for the infimum-stability for convex functions $f$ defined on $\R$ was given by Theorem 1 of A Necessary and Sufficient Condition on the Stability of the Infimum of Convex Functions (one can also use the arXiv version).

The proof of the just mentioned result takes about 4 pages. No extension of that result to $m>1$ seems to be known.