If $F$ is strictly increasing with closed image, then $F$ is continuous Let $F$ be a strictly increasing function on $S$, a subset of the real line. If you know that $F(S)$ is closed, prove that $F$ is continuous.

Let $f$ be any strictly increasing (not necessarily strictly) function on $S$.  To show that $f$ is continuous on $S$, it is enough to show that it is continuous at $x$ for every $x \in S$.  If $x$ is an isolated point of $S$, every function is continuous at $x$, so assume otherwise.
The key here is that monotone functions can only be discontinuous in a very particular, and simple, way.  Namely, the one-sided limits $f(x-)$ and $f(x+)$ always exist (or rather, the first exists when $x$ is not left-isolated and the second exists when $x$ is not right-isolated): it is easy to see for instance that 
$f(x-) = \sup_{y < x, \ y \in S} f(y)$.
Therefore a discontinuity occurs when $f(x-) \neq f(x)$ or $f(x+) \neq f(x)$.  In the first case we have that for all $y < x$, $f(y) < f(x-)$ and for all $y \geq x$, $f(y) > f(x-)$.  Therefore $f(x-)$ is not in $f(S)$.  But by the above expression for $f(x-)$, it is certainly a limit point of $f(S)$.  So $f(S)$ is not closed.  The other case is similar.
Other nice, related properties of monotone functions include: a monotone function has at most countably many points of discontinuity and a monotone function is a regulated function in the sense of Dieudonné.  In particular the theoretical aspects of integration are especially simple for such functions.  
Added: As Myke notes in the comments below, the conclusion need not be true if $f$ 
is merely increasing (i.e., $x_1 \leq x_2$ implies $f(x_1) \leq f(x_2)$).  A counterexample 
is given by the characteristic function of $[0,\infty)$.
