I want to prove :
Every open subset $U$ of a locally connected space $X$ is locally connected.
There is already a topic (Show that every open subset of a locally connected space is locally connected.) that deals with this question, but for some reason, I can't understand what they are actually saying there (maybe because we have different definitions of "local connectedness").
To me, a space $X$ is locally connected means : every point of $X$ has a connected neighborhood.
Here is my try. Let $x \in U$. Since $X$ is locally connected, there exists $C$ connected subspace of $X$ and $V$ open in $X$ such that $x \in V \subset C$. So $x \in U \cap V \subset U \cap C \subset U$, where $U \cap V$ is an open set of $U$, but the problem is that I can't say $U \cap C$ is connected, as there are cases where the intersection of a connected set and an open set has infinitely many connected components.
Any help would be appreciated.
Edit :
Sorry, the actual definition of "local connectedness" is :
A space $X$ is locally connected if every point $x \in X$ has a neighborhood base consisting of connected subspaces.
Here is the proof of the aforementioned result.
Let $V$ an open subset of $U$ such that $x \in V \subset U$. Since $U$ is open in $X$, $V$ is open in $X$. Since $X$ is locally connected, there exists $x\in W \subset C \subset V$ where $W$ is open in $X$ and $C$ is connected. Note that $W$ is then open in $V$. So $C$ does the job.
