I have read about $\mathbf{CW}$-complexes in two different books. This is not a question about equivalence of axiomatic and inductive definitions. I have found two variants of axiomatic definition. More specifically, I talk about $\mathbf{C}$-axiom. First variant (say $\mathbf{C_1}$): for each cell $e$ in cell-complex $X$ the closure $\bar{e}$ is contained in some finite (disjoint) union of cells. Second variant (say $\mathbf{C_2}$): for each cell $e$ the subcomplex $X(e)$ is finite, - where $X(e)$ is the intersection of all subcomplexes, which contains the cell $e$.
So, I want to know if axioms $\{\mathbf{(C_1),(W)}\}$ are equivalent to axioms $\{\mathbf{(C_2), (W)}\}$?
It is clear that the second one is not weaker then the first. So, I see, how to deduce the first from the second. If we assume the first, then I can prove that $X(e)$ is closed in X (like all subcomplexes). But I don't see that it must be finite. On the other hand, the closure $\bar{e}$, while is contained in finite union of cells, doesn't must be a subcomplex.
To prove that $\{\mathbf{(C_1),(W)}\}$ implies $\{\mathbf{(C_2), (W)}\}$, use induction on the dimension $n$ of the cell $e$ to prove that $\bar e$ is contained in a finite subcomplex.
If $e$ is 0-dimensional then $e$ is a subcomplex all on its own.
Suppose by induction that $e$ is an $n$-cell with $n \ge 1$, and that the closure of each cell of dimension $\le n-1$ is contained in a finite subcomplex. We know by $\mathbf{(C_1)}$ that $\bar e$ is contained in a finite union of cells. It follows $\bar e - e$ is contained in a finite union of cells. Furthermore since $\bar e - e$ is contained in the $n-1$ skeleton, it follows that $\bar e - e$ is contained in a finite union of cells of dimension $\le n-1$. Each of the latter cells is, by induction, contained in a finite subcomplex, so the union of those subcomplexes is a finite subcomplex that contains $\bar e - e$. By intersecting with the $n-1$ skeleton, we obtain a finite subcomplex of the $n-1$ skeleton that contains $\bar e - e$. And now, adding in the cell $e$, we obtain a finite subcomplex that contains $\bar e$.