homeomorphisms preserve connected components

Let $X,Y$ be topological spaces and $X=\bigcup X_i$, $Y=\bigcup Y_j$ be decompositions into connected components.

Proposition. Assume that $f:X\to Y$ is a homeomorphism. Then for any $i$ there exists $j$ such that $f(X_i)=Y_j$.

Proof. Take any $i$. Because $f$ is continuous $f(X_i)$ is connected, then there exists $j$ such that $f(X_i)\subseteq Y_j$ (because $Y_j$ is a connected component). Now $f$ is a homeomorphism, $f^{-1}(Y_j)\cap X_i\ne \mathrm{\varnothing }$, $Y_j$ is connected and $X_i$ is a connected component, so $f^{-1}(Y_j)\subseteq X_i$. Thus $Y_j\subseteq f(X_i)$, which completes the proof. $\mathrm{\square }$