Definition of separated:
$E$ is separated if $A, B \neq \varnothing$, $A \cup B = E$, and there are not convergent sequences in $A$ that have limit points in $B$, and vice versa.
Definition of convergent sequence:
$x_n \to x$ if $\forall \epsilon > 0$, $\exists N$ such that $\forall n \geq N$, $d(x_n, x) < \epsilon$.
It's pretty intuitive that the intervals $[0, \frac{\pi}{2})$ and $(\frac{3\pi}{4}, 2\pi)$ are separated, but it's not intuitive to me that there aren't any sequences in one interval that can't converge to a limit point in the other interval. But when I think about convergent sequence geometrically, it means that if I have some small neighborhood about the limit point, it contains terms in the sequence, or something like that. However I don't know how to rigorize this notion in terms of an official proof.
Assume by contradiction there is a sequence $(a_n)_{n\in\mathbb{N}}$ in $A=[0,\frac{\pi}{2})$ converging to a limit $a\in B=(\frac{3\pi}{4},2\pi)$. In particular, for $$\varepsilon\stackrel{\rm def}{=} \frac{\frac{3\pi}{4}-\frac{\pi}{2}}{4}$$ there exists $N_\varepsilon$ such that for any $n\geq N_\varepsilon$, $\lvert a_n - a\rvert \leq \varepsilon$. But as $a\in B$ and $a_n \in A$, we have for all $n\geq 0$ $$ \lvert a_n - a\rvert \geq \inf_{x\in A, y\in B} \lvert x-y \rvert = \frac{3\pi}{4} - \frac{\pi}{2} > \varepsilon $$ leading to a contradiction. (The other direction -- limits in $A$ of sequences in $B$ -- is similar)