Is every disconnected closed set the disjoint union of two nonempty closed sets?

Question

In section 1.2 of Ash and Novinger's Complex Variables, it is stated that if $X$ is a metric space and $S\subseteq X$ is a disconnected closed set, then $S$ can be written as a disjoint union of two nonempty closed sets.

I have trouble to see why this is true. After reading some proofs for the similar statement that "a topological space is disconnected iff it is a disjoint union of two nonempty closed sets", I managed to produce the following proof:

1. Suppose $S$ is disconnected. Then there exist two disjoint open sets $A,B\subset X$ such that $S\cap A$ and $S\cap B$ are nonempty and $S=(S\cap A)\cup(S\cap B)$.
2. Thus $S=A'\cup B'$, where $A'=S\cap A$ and $B'=S\cap B$ are two disjoint nonempty open sets in the subspace topology of $S$.
3. Hence $A'=S-B'$ and $B'=S-A'$ are also closed in the subspace topology of $S$.
4. Since $A'$ and $B'$ are closed in $S$ and $S$ is closed in $X$, the two sets $A'$ and $B'$ must be closed in $X$.
5. Now $S=A'\cup B'$ is a disjoint union of two nonempty closed sets in $X$.

Apparently, we don't need $X$ to be a metric space. I have several questions:

• Is my proof correct?
• Are there simpler proofs (without assuming that $X$ is endowed with a metric)?
• Is the converse true for a general topological space $X$?

Answer 1

$S$ is disconnected so it can be written as $A \cup B$ where $A$ and $B$ are disjoint,non-empty and open in $S$ (general definition of a connected space applied to $S$).

As $A$ and $B$ are open in $S$, and they are each other's complement (in $S$), they are both also closed in $S$. And a closed subspace of a closed subspace is closed in the ambient space (always). So $A$ and $B$ are also disjoint non-empty, closed subsets of $X$ with union $S$.