Demonstration: Let $x, y \in X$ be different. Because $X$ is completely regular, there is $f: X \to [0,1]$ continuous such that $f (x) = 0$ and $f (y) = 1$. Since $X$ is connected, $f [X]$ is an interval in $\Bbb R$, but $1,0 \in f [X]$, so $f [X] = [0,1]$.
I found this corollary, but I am unable to understand the proof. Can anyone help me with this? Thanks in advance.
Suppose there is some $a \in (0,1)$ that is not in $f[X]$. Then $X = f^{-1}[[0,a)] \cup f^{-1}[(a,1]]$ (if $x \in X$, then $f(x) \neq a$ by assumption so either $x$ is in the left hand or the right hand set). Both of these sets are open by continuity of $f$ and they are disjoint and non-empty ($x$, which maps to $0$ is in the left set, $y$ in the right). These facts combined imply that $X$ is disconnected which is a contradiction. It follows that $f[X]= [0,1]$.
The argument as given is summarising this argument by using two (presumably earlier) theorems: the continuous image of a connected set is connected, and a connected subset of $\Bbb R$ (or any space in an order topology) is order convex.