Hatcher explains on page 5 how a CW complex can be constructed inductively by attaching $n$-cells i.e. open $n$-dimensional disks.
On page 520 in the appendix he writes "A finite CW complex, ... , is compact since attaching a single cell preserves compactness."
Now my question: why is this obvious? An open disk is not compact, so how can I see that sticking two together is?
Many thanks for your help!
You are attaching closed discs in a CW complex (In the notation of hatcher $D^n$ is the closed $n$-disc cf. page XII). Each closed disc is compact.
The CW complex is formed by taking a quotient of a compact space - the finite union of compacts is compact. Taking a quotient preserves compactness since the quotient space is the image of the original space under the projection map, and continuous maps preserve compactness.