Perhaps the fastest proof of uncountability of is using the diagonalization trick. However, there’s another proof that I learnt from Topology by Munkres (Theorem 27.7 in page 176) which is nice in its own way.
It is enough to show that the closed interval is uncountable. Note that has the following three properties:
(Definition: If is a topological space, then is called an isolated point of if the singleton is an open set in ). And now the general theorem follows:
Theorem. If is a non-empty compact Hausdorff topological space having no isolated points, then is uncountable.
Proof. (Following Munkres). The proof proceeds in two steps. The first step will use Hausdorff condition and the non-existence of isolated points. The second step will use compactness (via the equivalent formulation stating that if any family of closed sets in satisfies the finite intersection property, then ).
Step 1: We show that if and is any non-empty open set in then there exists a subset such that . First we choose such that (this is possible: if , then let be any element of ; if then because is not open but is open, hence we can pick different from ). Next using Hausdorff condition, we can find disjoint neighbourhoods and of and respectively. It follows that satisfies (Why? If , then every neighbourhood of would intersect . But is a neighbourhood of , and , and so , contradiction).
Step 2. We will show that any function is not surjective. This will establish uncountability of . Let . Applying Step 1 to and , we get a subset such that . We inductively define as follows. If is already determined, we apply Step 1 to and to get a subset such that . So we have a nested sequence of non-empty closed sets:
Since is compact, the intersection is not empty. If , then for each (otherwise, which is a contradiction). Thus, is not surjective.