I'm having trouble with the following part in the proof of the stability theorem on Pg 36 of Guillemin and Pollack's Differential Topology: They write that since $X$ is compact, it follows that any open neighbourhood of $X\times \lbrace 0 \rbrace$ in $X \times [0,1]$ contains $X \times [0,\epsilon]$, for a suitable $\epsilon$. I cannot see how $X$'s compactness has anything to do with this: this depends only on the local compactness of the closed interval.
But, compactness is an essential condition for the theorem and this is the only point where it enters the proof (except part(e) of the theorem). Any help will be greatly appreciated.
Added I have a feeling that they want $X\times[o,\epsilon]$ to be compact, then prove something for open sets indexed by all points in this set (i.e. the obvious open cover) and then pass onto a finite cover...
Added I think I was reading the proof wrong. I am just writing this in case someone makes a similar silly mistake. My line of thought was that any open nbd of the slice $X\times\lbrace 0 \rbrace$ in $X \times I$ is of the form $X \times G$ where $G$ is open in $I$, since $I$ is locally compact, we can find a $[0,\epsilon]$ inside $G$ and $X\times [0,\epsilon]$ will cover $X\times\lbrace 0 \rbrace$. But what they want to do in the proof is to prove something locally at a point on $X\times \lbrace 0 \rbrace$ and boost it to the whole thing.