Exercise Prove: Let X be connected, and C a connected subset, and Q either open-and-closed, or a component, or a quasicomponent in X\C. Then X\Q is connected.
Proof Let us use + here to denote a separation: a decomposition into the union of two open-and-closed subsets. A disconnected space has a separation where both parts are nonempty.
(i) Let Q be nonempty and open-and-closed in X\C, say X\C = Q + R. If X\Q = Y + Z, where C is contained in Y, then Z is open-and-closed in X, hence empty.
(ii) Let Q be a component of X\C. If X\Q = Y + Z, where C is contained in Y, then by (i) X\Y is connected, contains Q and is contained in X\C. Hence X\Y = Q and Z is empty.
(iii) Let Q be a quasicomponent of X\C. Since Q is the intersection of all open-and-closed subsets S of X\C containing it, X\Q is the union of all subsets X\S. Each X\S contains C and is connected by (i), so their union is connected as well.