Suppose we are given a countable chain of topological spaces $X_0 \subset X_1 \subset X_2 \subset \cdots$ and let $X = \bigcup_n X_n$; and suppose further that for each $n$ we have a deformation retraction $F_n : X_{n+1} \times I \to X_n$. I'd like to build a deformation retraction from $X$ to $X_0$ by performing $F_n$ in the time interval $[1/2^{n+1}, 1/2^n]$, holding each point of $X_{n+1} - X_n$ stationary outside of this interval.
I'm having some trouble showing that this map is continuous. We can get continuity on $X \times (0,1]$ easily from the pasting lemma, but I don't know how to enlarge this to all of $X \times I$, due to the weird behavior of the function at the start of the interval.
EDIT: Just learned the map isn’t continuous in general, so let $X$ be a CW complex and the $X_n$’s the associated skeleta.
It's not true in general, so you're going to have to figure out what extra hypotheses are needed for the proof and are true in whatever application you have in mind.
For a simple counterexample, take $$X = S^1 = \{(\cos(2 \pi \theta),\sin(2 \pi \theta) \mid \theta \in (0,1]\} \subset \mathbb R^2 $$ with the subspace topology. And then take $$X_n = \{(\cos(2 \pi \theta), \sin(2 \pi \theta) \mid \theta \in (1/n,1] \} \subset X $$ also with the subspace topology. Each $X_n$ deformation retracts to $(1,0)$, but $S^1$ does not deformation retract to $(1,0)$.
I'll throw out one interesting and broad situation where it does work in general, namely where $X$ is a CW complex. The CW topology can be used to show that the continuous extension to $X \times [0,1]$ exists.