What is enough to conclude that something is a CW complex? This question was something I considered when looking into CW-structures on Grassmannians, but I found no general treatment of this in the literature:
Question: Assume that $X$ is an $n-1$ dimensional finite CW complex. and assume that $X'$ is given as a a set by the disjoint union of $X$ and a single open cell $e$ of dimension $n$. I.e. $e$ is an open subspace of $X'$ homeomorphic to the open $n$ disc (and of course $X$ is homeomorphic to the complement). Assume also that $X'$ is compact Hausdorff. Is $X'$ homeomorphic to a CW complex given by attaching a single $n$ cell to $X$?
Remark: Maybe I am missing some obvious counter example!
 A: Let $X$ be the closed subspace of $R^2$ which is the union of $0\times [-1,1]$ and the point $(1, \sin(1))$.  Let $e$ be the graph in the plane of $f(x)=\sin(1/x)$ for $x\in (0,1)$ (the "topologists sine curve"), and let $X'=X\cup e$, 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 $\Phi\colon[0,1]\to X'$ 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 $f\colon (S^2-\{p\})\to R$ given by $f(x)=\sin(1/|x-p|)$.  Let $X'$ be the closure of the graph of $f$ inside $S^2\times [-1,1]$; 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. 
