Some conditions for continuity of a real function Let $f:S\to \mathbb{R}$ be strictly monotone where $S\subseteq \mathbb{R}$. If the one of following conditions holds
$f(S)$ is open or $f(S)$ is closed, or $f(S)$ is connected,
then $f$ is continuous.
Why?
 A: First off, we are going to state an important theorem relative to discontinuities of monotone functions. If I remember correctly, there is a proof in Walter Rudin's book (the baby one).

Monotone function only have jump or removable singularities.

Thus, if your function $f:S\to \mathbb{R}$ is discontinuous, it may only have these types of singularities. Let's suppose $f$ and discontinuous at a point $x\in S$. We will show that $f(S)$ is neither closed nor open and not connected. Thus, by contraposition, your statement will be true.
Suppose $x\in \mathrm{int}\; S$. Then, we have 
$$\lim_{y\to x^-} f(y)\leq f(x)\leq \lim_{y\to x^+} f(y)$$
If equality hold for one case (without any loss of generality $\lim_{y\to x^-} f(y)= f(x)$), then the image $f(S)$ is not closed nor open and disconnected. In fact, the complement of the region close to $f(x)$ looks like the interval $(a,b\rbrack$. If equality doesn't hold, then the region of $S$ near $f(x)$ is just a single point. The complement of $f(S)$ looks like $\lbrack a,b)\cup (b,c\rbrack$ in this region. You should fill in the details here.
If $x\not\in \mathrm{int}\; S$, again, you can do the same kind of reasonning with limits.
I think this shows the idea. I do not claim this to be a complete proof. The details should follow easily from here fortunately.
