No. For instance, let $Y$ be any space that is not locally connected, and let $X=Y\cup\{x\}$ where $x$ is a new point and a subset of $X$ is open iff either it is empty or it has the form $U\cup\{x\}$ where $U\subseteq Y$ is open. Then any two nonempty open subsets of $X$ intersect, so $X$ is trivially locally connected. But $X\setminus\{x\}=Y$ is not locally connected.
It is true if $X\setminus\{x\}$ is open in $X$ (in particular, if $X$ is $T_1$), since any open subspace of a locally connected space is locally connected (you can just restrict your connected local bases at each point to the sets that are contained in the subspace).