gn.general topology - What is enough to conclude that something is a CW complex?

Let $X$ be the closed subspace of $R^2$ which is the union of $0times [-1,1]$ and the point $(1, sin(1))$. Let $e$ be the graph in the plane of $f(x)=sin(1/x)$ for $xin (0,1)$ (the "topologists sine curve"), and let $X'=Xcup e$, viewed as a subspace of $R^2$.

I believe that $X'$ is closed, and so is compact Hausdorff, $e$ is an open subset of $X'$ homeomorphic to the open interval, and $X$ is a CW-complex. But $X'$ is not obtained by attaching a $1$-cell to $X$: you can't produce a map $Phicolon[0,1]to X'$ which restricts to a homeomorphism $(0,1)to e$.

This is not quite a counterexample to your problem, because you wanted the new cell $e$ to have greater dimension than the cells of $X$. But it seems like you could use this kind of idea to make a counterexample.

Added. Let $S^2$ be the unit sphere in $R^3$, pick a point $p$ on $S^2$, and consider the function $fcolon (S^2-{p})to R$ given by $f(x)=sin(1/|x-p|)$. Let $X'$ be the closure of the graph of $f$ inside $S^2times [-1,1]$; this should consist of $X={p}times [-1,1]$ (a CW complex) and $e=$ graph of $f$ (homeomorphic to an open $2$-disk). This would seem to be the counterexample you want.