Let $f:X\to Y$; let $X$ and $Y$ be metrizable with metrics $d_X$ and $d_Y$, respectively. Then continuity of $f$ is equivalent to the requirement that given $x\in X$ and given $\epsilon>0$, there exists $\delta>0$ such that $d_X(x,y)<\delta\implies d_Y(f(x),f(y))<\epsilon$.
I know the following result: A mapping $f$ of a metric space $X$ into a metric space $Y$ is continuous on $X$ if and only if $f^{-1}(V)$ is open $X$ for every open set in $Y$. This is theorem 4.8 of Baby Rudin. Using this result, theorem 21.1 is straightforward. Here’s a proof Exercise 1, Section 18 of Munkres’ Topology. In that post I proved notion of open set in metric space and topological space are same. Do I need to rewrite the proof or above argument is enough? Even if I rewrite the proof, idea in both(thm 4.8 & thm 21.1) proof are similar. In fact, in lemma 21.2, proof is almost identical to theorem 3.2(d) of Baby Rudin or https://math.stackexchange.com/a/4338980/861687.
Best Answer
This theorem is best proved by using that $(X,d_X)$ induces a topological space $(X,\mathcal{T}_{d_X})$ with all metric open balls as its base and $(Y,d)$ likewise induces a space $(Y,\mathcal{T}_{d_Y})$ and for topological spaces we have the open set definition of continuity (inverse images of open sets are open).
So let $f$ be topology-continuous. Let $x \in X$ and $\varepsilon>0$. By definition $O:=B_{d_Y}(f(x),\varepsilon)$ is open in $Y$ and $x \in f^{-1}[O]$ which is an open set by said continuity. So there is some $\delta>0$ so that $x \in B_{d_X}(x,\delta) \subseteq f^{-1}[O]$ as open balls form a base for $X$ and this $\delta>0$ is as required: $$d_X(x,x') < \delta \implies x' \in B_{d_X}(x,\delta) \implies x' \in f^{-1}[O] \implies f(x') \in O \\ \implies d_Y(f(x'),f(x)) < \varepsilon$$ as required. So $f$ is epsilon-delta continuous.
The converse is very similar and is left to you. It essentially uses the same ideas/steps to show that $f^{-1}[O]$ is open whenever $O$ is.