Let $X$ be a cell complex with one $0$-cell and an arbitrary (possibly infinite of any degree) number of $1$-cells which are self-loops to the only $0$-cell. Clearly $X$ is closure-finite (property C in CW). Is the topology of $X$ necessarily coherent with the set of closed cells of $X$ (property W in CW)? That is, is $X$ a CW complex?
Note: If $X$ is first countable, then I can show that being locally finite is equivalent to property W. This result can be used to show, for example, that the closed infinite (countable or uncountable) broom does not have property W. There is a question here which states that an infinite bouquet of circles is not first countable, which excludes using this result.
Best Answer
Since you haven't specified the topology on $X$, the answer is no in general. For example, $X$ could be the union of the circles $C_n$ in the Euclidean plane with center at $(1/n,0)$ and radius $1/n$. If we topologize this $X$ as a subspace of the plane, it consists of a single $0$-cell $O$ (the origin) and infinitely many loops $C_n$ at $O$. But it's not a CW complex; it doesn't have the weak topology. For example, the set consisting of the closed right halves of all the $C_n$'s (i.e. the part of each $C_n$ where $x\geq 1/(2n)$) meets each $C_n$ in a closed set, but its closure contains $O$.