Equivalence of locally connected?

Question

Is true this equivalence?

$A\subseteq X$ is locally connected in $X$ if and only if for each $U$ open in $X$ such that $x\in U$, exists a open connected set $K$ such that $x\in K \subseteq U$ and $K \cap A$ is connected.

$(\Leftarrow)$ Is obviously, but the converse I can´t. Any hint?, thanks.

Answer 1

$\Rightarrow$ is false. Let $X = \{0\} \cup \bigcup_{n=1}^\infty \{1/n\}$ with the topology inherited from $\mathbb R$. The subspace $A= \{0\}$ is locally connected, but $0$ has no open connected neighborhood $K$ in $X$.