I would like to show the following implication.
Let $f$ be a function from $\mathbb{R}^n$ to $\mathbb{R}^m$. If $f^{-1}(U)\subset\mathbb{R}^n$ is open for every open $U\subset\mathbb{R}^m$, then $f$ is continuous.
Here is my attempt:
Let $B$ be the $m$-ball of radius $\epsilon>0$ centered at $f(a)$, where $a$ is a point in the domain of $f$. By hypothesis, $f^{-1}(B)$ is open. Thus, there is an $n$-ball of radius $\delta$ for some $\delta>0$ centered at $a$ that lies entirely within $f^{-1}(B)$. Call this $n$-ball $B^{-1}$. Note, $f(B^{-1})=B$ is not necessarily true, as $f$ isn't necessarily surjective.
But we know that, by the definition of $B$, for every point $x$ within $f^{-1}(B)$, we have that $\left|f(x)-f(a)\right|<\epsilon$. We also know that $B^{-1}\subset f^{-1}(B)$. Thus, we can say that there exists a $\delta>0$ such that $\left|f(x)-f(a)\right|<\epsilon$ whenever $0<\left|x-a\right|<\delta$.
Is this fine? I have tried not to look at any other proofs of this either here on MSE or anywhere else.
You're very close. In particular, there is an open $n$-ball centered at $a$ contained entirely in $f^{-1}(B),$ having radius $\delta$ for some $\delta>0.$
You should probably then use the definition of preimage to finish the proof.