Let $X$ be a compact Hausdorff space without any isolated point. Show that $X$ is uncountable.
As $X$ is compact Hausdorff, it is normal. then for any two distinct points $x$ and $y$, we have a continuous map $f$ from $X$ to $[0,1]$ such that $f(x)=0$ and $f(y)=1.$
As $X$ is compact, $f(X)$ is closed and bounded in $[0,1]$. Now I want to show that $f(X)$ contains an interval by using the fact that $X$ has no isolated point. Some one Help me.
One way to do this is via a version of the Baire Category Theorem, which states that every locally compact Hausdorff space is a Baire space, i.e. it is not the union of a countable set of nowhere dense subsets. Recall that compact implies locally compact, so $X$ is not the union of a countable set of nowhere dense subsets. But $$X=\bigcup_{x\in X}\{x\}$$ and each $\{x\}$ is nowhere dense since $x$ is not an isolated point, thus $X$ must be uncountable.
The idea of using normality to show that $X$ can be mapped onto the interval $[0,1]$ is a good one, but it doesn’t work: if $X=A\cup B$, where $A$ and $B$ are disjoint clopen (= closed and open) sets, your function will be
$$f:X\to[0,1]:x\mapsto\begin{cases}0,&\text{if }x\in A\\1,&\text{if }x\in B\end{cases}$$
or the opposite one that takes $A$ to $1$ and $B$ to $0$: it will assume only the two values $0$ and $1$ and therefore will show only that $|X|\ge 2$, which doesn’t help.
This answer to a slightly different question uses an idea that can be modified to work here. In fact, it’s simpler in this setting. The basic idea is to start with non-empty open sets $U_0$ and $U_1$ whose closures are disjoint. Inside $U_0$ find non-empty open sets $U_{00}$ and $U_{01}$ whose closures are disjoint, and inside $U_1$ find non-empty open sets $U_{10}$ and $U_{11}$ whose closures are disjoint. In general, if $\sigma$ is any finite sequence of zeroes and ones, inside $U_\sigma$ find non-empty open sets $U_{\sigma^\frown 0}$ and $U_{\sigma^\frown 1}$ whose closures are disjoint. (Here $\sigma^\frown i$ is the finite sequence obtained by appending $i$ to $\sigma$.) The hypotheses on $X$ ensure that this is possible; why?
Now show that if $\sigma=\langle i_k:k\in\Bbb N\rangle$ is any infinite sequence of zeroes and ones, then
$$\bigcap_{k\in\Bbb N}\operatorname{cl}U_{i_0i_1\dots i_k}\ne\varnothing\;,$$
and use this to show that $|X|\ge 2^\omega=\mathfrak{c}$.
You may find that it helps your intuition to think of the sets $\operatorname{cl}U_\sigma$ as being like the closed intervals in the construction of the middle-thirds Cantor set: $\operatorname{cl}U_0$ is like $\left[0,\frac13\right]$, $\operatorname{cl}U_1$ is like $\left[\frac23,1\right]$, $\operatorname{cl}U_{01}$ is like $\left[\frac29,\frac13\right]$, and so on.
