Proposition2.2 (ncatlab): Suppose $\phi: X \to Y$ is a simplicial approximation to a map $f: {|X|} \to {|Y|}$, and suppose $f = {|\phi|}$ on a subset $A \subseteq {|X|}$. Then there is a homotopy ${|\phi|} \simeq f \; rel \; A$.
Proof Define the homotopy $H$ by $H(\alpha, t) = t f(\alpha) + (1-t){|\phi|}(\alpha)$. This makes sense since if $f(\alpha) \in {|s|}$, then ${|\phi|}(\alpha) \in {|s|}$ and ${|s|}$ is closed under convex combinations. Clearly $H(\alpha, t) = f(\alpha)$ for all $t$ on a set where $f$ and ${|\phi|}$ agree.
I see why if $|Y| \subseteq \mathbb{R}^N$ the homotopy $H$ is continuous as $\mathbb{R}^N$ is a topological vector space. But this need not be the case in general? What if $Y$ is an infinite dimensional simplicial complex?
To check $H$ is continuous, you only need to check that it is continuous on $\sigma\times I$ for each finite simplex $\sigma\subseteq |X|$. But any such $\sigma\times I$ is compact, and so the images of $f$ and $|\phi|$ on $\sigma$ are contained in a finite subcomplex of $|Y|$. The image of $H$ on $\sigma\times I$ will then be contained in that finite subcomplex, so to check its continuity you may assume $Y$ is finite.