A space $X$ is locally connected if and only if every component of every open set of $X$ is open?

If $X$ is locally connected and $C$ is a connected component of an open subset $U \subseteq X$, then every point $c \in C$ contains a connected neighborhood which lies in $U$ (because $X$ is locally connected and $U$ is open). But then it has to lie completely in the connected component of $c$, i.e. in $C$. This shows that $C$ is open.

The proof of the converse is very, very similar. Can you write it down for yourself?

Suppose $X$ is locally connected, and take an open set $A$ and a component $C$ of $A$. We want to show that $A \backslash C$ is closed in $A$. Take a point $x$ in the closure of $A \backslash C$, by local connectedness we can find a connected neighborhood $N$ of $x$. If $x \in C$ then $N \subseteq C$, but $N$ intersects $A \backslash C$, which is absurd. So $C$ is open in $A$, hence in $X$.

The converse follows from the obvious fact that every open set is the union of its components.

As for your second question, the statement is true, but it follows directly from the definition.