Clearly reflexivity and symmetry hold. Assume that $x \sim y$ and $y \sim z$, but $x \nsim z$. This would imply that there is a separation $X = A \cup B$ such that $x \in A$ and $z \in B$. Since the separation is equal to the whole set, it must be true that $y \in A$ or $y \in B$. In the former case, this would contradict $y \sim z$, while the latter would contradict $x \sim y$. Thus, it must be true that $x \sim z$. Thus, transitivity holds, and the relation is an equivalence relation.

Let us consider the component $C$, determined by some reference point $x$. Let $C'$ be the quasicomponent determined by $x$. Let us pick some $y \in C$. By definition, it must be true that $x$ and $y$ lie in a connected subspace $L$ of $X$. Assume that $y \notin C'$. It would then hold true that there exists a separation $X = A \cup B$ such that $x \in A$ and $y \in B$. If this was the case, then $L \cap A$ and $L \cap B$ would be disjoint open sets of $L$, whose union is $L$. This contradicts the fact that $L$ is connected. Thus, it must be true that $y \in C'$. It follows that $C \subset C'$ and each component is contained in a quasicomponent.

Consider the quasicomponent $C'$ determined by the reference point $x$. Also consider the component $C$ determined by $x$. Let us have $y \in C'$. This implies there exists no separation with $x$ and $y$ in different sets. Assume that $y \notin C$. This implies that $y \in X - C$. We know that components are closed. It follows that $X - C$ is open. We also know that if $X$ is locally connected, then every component will be open. Thus, $C$ is open. This means that $C$ and $X - C$ are two disjoint open sets whose union is $X$, with $x \in C$ and $y \in X - C$. This contradicts the fact that $x \sim y$. It follows that $y \in C$. Combinoting this with the previous lemma, and we have $C' = C$.