This problem is from Andreas Gathmann's lecture notes on Topology.
Similar questions have been asked already.
The problem:
Let $X$ be a top. space.
Show that:
$X$ is connected $\iff$
for every continuous function
$f:X \rightarrow \mathbb{R}$ and any
two points $x, y \in X$ every value
between $f(x)$ and $f(y)$ is attained
by $f$ on $X$.
$\Rightarrow:$
Let $X$ be connected, then $f(X)$ is connected since $f$ is continuous.
Let $x, y \in X$, and $f(x) <f(y)$.
Assume there is a $z \in \mathbb{R} $ with
$f(x) <z<f(y)$ and $z \not \in f(X) $.
Then $U:=f(X) \cap (-\infty, z)$, $(\not = \emptyset)$, and
$V:=f(X) \cap (z, \infty)$,
$(\not =\emptyset)$,
are open in $f(X)$ (subspace topology), and $U \cap V =\emptyset$.
We have
$f(X) =U \cup V$, a contradiction, since $f(X)$ is connected.
$\Leftarrow$:
Assume $X$ is not connected.
Then there exist open and disjoint sets $U$ and $V$(both not empty) s.t.
$X=U \cup V$.
Consider
$f:X \rightarrow \mathbb{R}$ where
$x \rightarrow 0$ for $x \in U$,
$x \rightarrow 1$ for $x \in V$.
Then $f$ is continuous, however the Intermediate Value theorem is not applicable, and we are done.
Please correct and comment. Thank you.
Best Answer
Both proofs are correct, the first could be made shorter by not reproving known facts. If $f(x) < f(y)$ are in $f[X]$, then $[f(x), f(y)]$ is connected in $\Bbb R$ and so as $f[X]$ is connected, $[f(x), f(y)] \subseteq f[X]$ and you're done right away (your $z \in [f(x), f(y)]$ etc.). This is assuming you know the characterisation of connected subsets of $\Bbb R$, of course.
The reverse implication is quite a standard idea (using a continuous two-valued map), I see nothing to simplify it.