Comb space

Counterexample

Let $K = \set{0} \cup \set{1/n : n \ge 1} \subseteq \R$. The comb space is the subspace of $\R^2$ $$C = ([0, 1] \times \set{0}) \cup (K \times [0, 1]).$$ It consists of a base segment, the tooth $\set{0} \times [0, 1]$, and the teeth $\set{1/n} \times [0, 1]$.

The space $C$ is path-connected. Each point can be joined to $(0, 0)$ by going down its tooth and then along the base.

The space $C$ is not locally connected. Take $p = (0, 1)$ and the open set $U = C \cap B(p, 1/2)$. For large $n$, the set $U$ contains a piece of the tooth $\set{1/n} \times [0, 1]$ near height $1$. Any open set $V$ with $p \in V \subseteq U$ contains such pieces, but $U$ contains no part of the base. The vertical line $x = c$ with $c$ strictly between two teeth splits $V$ into two disjoint open parts, one of them containing $p$. So no open neighbourhood of $p$ inside $U$ is connected.