Let $X$ be a topological space. I have already shown that
(i) If $Q$ is a quasi-component of $X$ and $Q$ is open, then $Q$ is connected
(ii) If $X$ has only finitely many quasi-components, then each quasi-component is connected.
Now I want to show that if $X$ is a locally connected, then each quasi-component of $X$ is connected.
NOTE: The definition of quasi-component I am using: The quasi-component of $x$ is the intersection of all clopen subsets of $X$ containing $x$. Please use this definition.
I have three apparent options. Assuming $X$ is locally connected, I could show
(a) $X$ has finitely many quasi-components
(b) Any given quasi-component of $X$ is open
(c) Any given quasi-component is connected.
Option $(a)$ doesn't seem viable and so I haven't given it much thought. I have spent quite a while trying to show that a given quasi-component is open and trying to show that a given quasi-component is connected but to no avail.
Since $X$ is locally connected, given $x \in X$ and $U$ open in $X$ with $x \in U$, there exists a connected open subset $V \subset U$ with $x \in V$. I was thinking that from here I could show that $V$ is contained in the quasi-component of $x$, call it $Q_x$, so that $Q_x$ would be open, but that didn't work. Then I tried to run into a contradiction by supposing that $V$ intersects two distinct quasi-components of $X$ and nothing happened.
Any suggestions? Thanks :)