I think the condition about "closure-finiteness" for CW-complex is superfluous. Let $e$ be a cell in a CW-space $X$. Then, there exists a characteristic map $f:D^m\rightarrow X$ which maps $int(D^m)$ homeomorphically into $e$. Since $D^m$ is compact and $X$ is Hausdorff, we have $f(D^m)=\bar e$. This implies that $\bar e$ is compact. Since cells cover $X$, there must exist a finite number of cells covering $\bar e$. Where did I go wrong?
Cells are not open subsets of $X$, so it isn't obvious that only finitely many of them must cover the compact set $\overline{e}$. It is possible to prove that a compact subset of a CW-complex is contained in only finitely many cells, but the proof uses the closure-finite axiom!
If you're still having trouble wrapping your head around this, here's the standard counterexample. Take $X$ to be a closed $2$-disk. Say that it has one $2$-cell (with $\Phi$ just the identity map) and a $0$-cell for every point in the boundary of the disk. This satisfies all the axioms except closure-finiteness, but the closure of the $2$-cell consists of uncountably many $0$-cells.