A Space with Connected Components that are not Open
# Example
Consider $\mathbb{Q}$ equipped with the Subspace Topology of Euclidean Space. Then no Connected Component of $\mathbb{Q}$ is Open.
# Proof
Singletons are always Connected. If $S \subset \mathbb{Q}$ s.t. $|S| \geq 2$, then there exist distinct Rational Numbers $p< q \in S$. By Irrationals are Dense in Reals, there exists an $r \not\in \mathbb{Q}$ so that $p < r < q$. Thus, if we take $U = (-\infty, r)$ and $V = (r, \infty)$,
- $p \in U \cap S, q \in V \cap S$, so both are non-trivial
- $U \cap V = \emptyset$
- $(U \cup V) \cap S = S$
Thus $U, V$ disconnect $S$ and $S$ is not Connected. So every singleton in $\mathbb{Q}$ is a Connected Component. None of them are Open because every Open $U \subset \mathbb{R}$ contains an Interval, which contains infinitely many Rational Numberss (because Rationals are Dense in the Reals). $\blacksquare$