Why is a finite CW complex compact?

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?

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.
• But on page 5 he writes "open $n$-disk" where he explains how to construct a CW complex in step (2). He is attaching open disks! – Rudy the Reindeer Apr 21 '11 at 10:25
• @Matt- So usually the open disk' will correspond to what's called a cell,' but if you look carefully the way you actually construct the CW-complex is to glue the cell in along its boundary... i.e. look at a map $f: S^{n-1} \rightarrow X^{n-1}$ and use that to add an $n$-cell. – Dylan Wilson Apr 21 '11 at 10:44