Let $X$ be a topological space; let us define $x \sim y$ if there is no separation $X = A \cup B$ of $X$ into disjoint open sets such that $x \in A$ and $y \in B$.
This relation is an equivalence relation; the equivalence classes are called the quasicomponents of $X$. This I can show.
Moreover, each connected component of $X$ lies in a quasicomponent of $X$. This too have I managed to show.
How to show that the components and quasicomponents of $X$ are the same if $X$ is locally connected?
Suppose that $X$ is locally connected.
Let $C$ be a component of $X$ and let $Q$ be the quasicomponent containing $C$. Suppose that $C \subset\neq Q$.
Let $D$ be the union of all the components of $X$ different from $C$ that intersect $Q$. Then $Q = C \cup D$.
And, $C \cap D = \emptyset$ because distinct components, being distinct equivalence classes of a certain equivalence relation, are non-empty and disjoint.
Now since $X$ is locally connected, components of open sets in $X$ are open.
Since $X$ itself is open, $C$ is open and $D$ is a union open sets and therefore open.
Thus we have written $Q$ as a union of two disjoint non-empty sets each of which is open in $X$.
Now let $x \in C$ and $y \in D$. Then can we obtain a separation $X = A \cup B$ of $X$ into disjoint non-empty open sets $A$ and $B$ such that $x \in A$ and $y \in B$?
Or, can we obtain a contradiction some other way?
Following the hint given in Stefan Hamcke's answer:
Suppose that $X$ is locally connected and suppose also that the component $C$ is properly contained in the quasicomponent $Q$.
Components of open sets are open in any locally connected topological space, and components of any topological space are closed.
Since $X$ is open, therefore $C$ is both open and closed in $X$.
Now suppose that $x \in C$ and $y \in Q - C$.
Then $X = C \cup (X - C)$ is a separation of $X$ into disjoint non-empty open subsets $C$ and $X - C$ such that $x \in C$ and $y \in X-C$. Then $x \not\sim y$, which contradicts the assumption that $x, y \in Q$ and $Q$ is a quasicomponent and hence $x \sim y$.
Thus our sopposition that $C$ is properly contained in $Q$ is wrong. So $C = Q$.
Is this proof correct now?