Let $f$ be a differentiable function on $\Bbb R$. Prove that $f$ is strictly increasing if and only if $f' \geq 0$ and the set $D = \{x : f'(x) = 0\}$ is totally disconnected.
Let $f$ is strictly increasing and let the set $D = \{x : f'(x) = 0\}$ is not totally disconnected. So one component of the set has at least two points, contradicting that $f$ is strictly increasing.
Is the logic correct?
You should explain better where the contradiction comes from. The main point is the Mean Value Theorem.
If $f$ is strictly increasing, then $f'(x)\geq 0$ by the Mean value theorem. If $x,y\in D$, say with $x<y$, the Mean Value Theorem again implies that there is a point $c\in (x,y)$ with $f'(c)>0$, so $x$ and $y$ do not belong to the same connected component of $D$, which is therefore totally disconnected.
Conversely, if $f'\geq 0$ and $D$ is totally disconnected, take $x<y$ in $\mathbb{R}$. The interval $(x,y)$ is not contained in $D$, otherwise it would be contained in a nontrivial connected component, so $f'(c)>0$ for some $c\in(x,y)$ and therefore $f(x)<f(y)$ (we don't need the mean value theorem here, just the fact that, since $f'(c)>0$, there exist $d,e$ sufficiently close to $c$ with $d<c<e$ and $f(d)<f(e)$).