Hatcher, p. 520 in the appendix:
A subcomplex of a CW complex $X$ is subspace $A \subset X$ which is a union of cells of $X$ such that the closure of each cell in $A$ is contained in $A$ ... It is easy to see by induction over the skeleta that subcomplex is a closed subspace.
I am wondering if I have the right idea of what this proof by induction should look like: suppose let $A,X$ be given as above. Then certainly, $A ~ \cap X^0$ is a closed subspace, as it is a discrete set. Then suppose $A ~ \cap X^{n-1}$ is a closed subspace. We have $$A ~ \cap X^n = (A ~\cap X^{n-1}) ~ \cup (\bigcup \bar{e_i})$$ Where the $e_i$ are the $n$-cells in $X^n$ which are also in $A$. The former set is closed by assumption. The latter must be also, because $X$ is closure-finite.* Hence $$A ~ \cap X^{n-1} \textrm{ is closed} \implies A~\cap X^n \textrm{ is closed}.$$ We are done.
*It is this step I am most unsure about. I guess my main question is whether or not the implication I use
$$X \text{ is closure-finite} \implies \bigcup \bar{e_i} \text{ is a closed subspace of }X$$
Is true. If not, then my proof does not work, and I'm left wondering what Hatcher's proof by induction is.
Your doubts are justified. Closure finiteness does not imply that $\bigcup \bar{e_i}$ is closed.
As an example take the CW-complex $X$ which is obtained from $X^1 = I = [0,1]$ by attaching $2$-cells $D^2_i$, $i \in \mathbb N$, via the constant maps $\phi_i : S^1 \to I, \phi_i(x) = 1/i$. Then $X^1 \cap \bigcup \bar{e_i} = \{1/i \mid i \in \mathbb N\}$ which is not closed in $X^1$.
Let $\{e^n_i\}_{i \in I_n}$ denote the set of open $n$-cells of $X$ and $\{e^n_i\}_{i \in I'_n}$ denote the subset of cells contained in $A$.
Let us now prove by induction that $A^n := A ~ \cap X^n = (A ~\cap X^{n-1}) ~ \cup \bigcup_{i \in I'_n} e_i^n$ is closed in $X^n$. The base case $n = 0$ has been treated in your question. Here is the induction step.
$X^n$ is obtained from $X^{n-1}$ by attaching $n$-cells via a collection of attaching maps $\phi_i :S^{n-1}_i \to X^{n-1}$. Thus there is a quotient map $q : X^{n-1} \sqcup \bigsqcup_{i \in I_n} D^n_i \to X^n$ which is the identity on $X^{n-1}$. Let $\Phi: \bigsqcup_{i \in I_n} D^n_\alpha \to X^n$ be the restriction of $q$.
Hence it suffices to show that $q^{-1}(A^n)$ is closed in $X^{n-1} \sqcup \bigsqcup_{i \in I_n} D^n_i$ which means
But 1. is the induction hypothesis because $A^n \cap X^{n-1} = A \cap X^{n-1}$. Concerning 2. we proceed as follows:
We have $\Phi^{-1}(A^n) = \Phi^{-1}(A^{n-1}) \cup \bigsqcup_{i \in I'_n} \mathring D^n_i$. We know that each $e_i$ with $i \in I'_n$ is an open $n$-cell of $X$ which belongs to $A$. Thus the boundary $\partial e_i$ of the closed $\overline{e_i} \subset A$ is contained in $A \cap X^{n-1} = A^{n-1}$. Hence $S^{n-1}_i \subset \Phi^{-1}(A^{n-1})$ and we conclude $$\Phi^{-1}(A^n) = \Phi^{-1}(A^{n-1}) \cup \bigsqcup_{i \in I'_n} D^n_i .$$ Clearly $\Phi^{-1}(A^{n-1})$ is closed in $\bigsqcup_{i \in I_n} D^n_i$ because $A^{n-1}$ is closed in $X^{n-1}$, hence also closed in $X^n$. But obviously also $\bigsqcup_{i \in I'_n} D^n_i$ is closed in $\bigsqcup_{i \in I_n} D^n_i$ which completes the proof.