The following is from Theorem 5.14 of John Lee's Introduction to Topological Manifolds.
Let $X$ be a CW complex. A subset of $X$ is compact if and only if it is closed in $X$ and contained in a finite subcomplex.
The proof for the only if direction is given as: suppose $K \subset X$ is compact. If $K$ intersects infinitely many cells, by choosing one point of $K$ in each such cell we obtain an infinite closed and discrete subset of $K$, which is impossible.
My question is why is this subset, say $A$, a closed subset of $K$? I think this is because $A'=\emptyset$, but I can't give a rigorous proof of this. I would greatly appreciate any help.
Let us prove a slightly more general theorem:
Let $A$ is a subset of $X$ such that $A \cap e$ is finite [this allows being empty] for each cell $e$ of $X$. Then $A$ is closed in $X$.
The set $A$ is closed iff $A \cap \bar e$ is closed in $\bar e$ for all cells $e$ of $X$. We can prove this by induction over the dimension of cells.
The base case $\dim e = 0$ is trivial.
Assume that $A \cap \bar e$ is closed in $\bar e$ for all cells $e$ of $X$ with $\dim e \le n$. This means that $A \cap X^n$ is closed in the $n$-skeleton $X^n$ and hence closed in $X$.
We have to show that $A \cap \bar e$ is closed in $\bar e$ for all cells $e$ of $X$ with $\dim e = n+1$.
Let $e$ be such a cell. We have $A \cap \bar e = (A \cap X^n) \cap \bar e \cup (A \cap e)$. Since $A \cap X^n$ is closed in $X$, $(A \cap X^n) \cap \bar e$ is closed in $\bar e$. But $A \cap e$ is finite, thus it is closed in $\bar e$. Since unions of closed sets are closed, we see that $A \cap \bar e$ is closed in $\bar e$.
By the way, the above argument also shows that $A$ is a discrete subspace of $X$. In fact, all subsets of $A$ are closed in $X$ and hence also closed in $A$. This means $A$ carries the discrete topology.