Show that for $x_o\in X$, the connected component of $x_o$ is connected.
Attempt: So I'm trying to show that assuming that the union of connected sets that contain $x_o$ is not connected results in a contradiction.
Let $x_o\in X$ and let $A_1$ and $A_2$ connected such that $x_o\in A_1,A_2$. Suppose $A_1\cup A_2$ is not connected. Then $\exists U,V$ open that disconnect $A_1\cup A_2$. Without loss of generality, let $U\cap A_1\neq\emptyset$. Then, $V\cap A_2\neq\emptyset$. If $V\cap A_1\neq\emptyset$, $U\cap A_2=\emptyset$. But then $U,V$ do not disconnect $A_1\cup A_2$, contradiction.