The conclusion of the theorem is that $X$ is uncountable. This is defined, usually, as "$X$ is not countable". And logically speaking (in first order logic) the way to prove a statement of the form $\lnot \phi$ is to assume $\phi$ and derive a contradiction from it (even Brouwer, in his intuition...

Suppose that $X$ is perfectly normal space. To show that $X$ is completely normal, I must show that every subspace of $X$ is normal. To that effect, let $Y$ be a subspace of $X$. Let $A, B$ be disjoint closed sets in Y. So there exist closed sets $A_1, B_1$ in X (these need not be disjoint) such ...

Assume there exists at least one $z\in E$ such that $z\rightarrow y$ for all $y\in E\setminus \{z\}$ and $\Bbb{P}_y(T_z<\infty)=1$ for all $y\in E\setminus \{z\}$. I need to show that the Markov chain $X$ is irreducible. The problem is I don't know where to start. So I know by definition that $...