Here we show as the hint suggested that a component of an open subset of $X$ is open. Let $U$ be open in $X$ and $C\subseteq U$ be its component. For any point $x\in C\subseteq U$ there is a connected subspace $S_x\subseteq X$ and an open neighborhood $V_x$ such that $x\in V_x\subseteq S_x\subseteq U$ . Since $S_x$ is a connected subset of $U$ , $S_x\subseteq C$ . Therefore, $C$ is the union of $V_x$ for all $x\in C$ and is open. Another way to prove this is to show that the space is locally connected at every point. This way might be better considering the next exercise (in the proof we should somehow use the fact that the space is weakly locally connected at every point, or at least at every point in some neighrborhood of $x$ , to prove that it is locally connected at $x$ ). Let $U$ be a neighborhood of $x$ . There is a connected subspace $S$ such that it contains a neighborhood of $x$ . Suppose there is a neighborhood of $x$ such that the space is weakly locally connected at every point in the neighborhood. The intersection of these two neighborhoods of $x$ is a neighborhood $V$ of $x$ such that it is contained in $S$ and the space is weakly locally connected at every point in the neighborhood. If $C$ is a component of $V$ containing $x$ then every point in $C$ is contained in a connected subspace that a) has to be contained in $C$ and b) contains a neighborhood of the point. Therefore, as before, we conclude that $C$ is open in $V$ , and, therefore, in $X$ , contains $x$ and is connected. I.e. the space is locally connected at $x$ . Note that for this to be true we need the space to be weakly locally connected not only at the point $x$ but at any point in some neighborhood of $x$ . The next exercise shows that weak local connectedness at the point only does not imply the local connectedness at the point.