**Proof.**
Let $\overline{T}$ be the closure of the irreducible subset $T$. If $\overline{T} = Z_1 \cup Z_2$ with $Z_ i \subset \overline{T}$ closed, then $T = (T\cap Z_1) \cup (T \cap Z_2)$ and hence $T$ equals one of the two, say $T = Z_1 \cap T$. Thus clearly $\overline{T} \subset Z_1$. This proves (1). Part (2) follows immediately from (1) and the definition of irreducible components.

Let $T \subset X$ be irreducible. Consider the set $A$ of irreducible subsets $T \subset T_\alpha \subset X$. Note that $A$ is nonempty since $T \in A$. There is a partial ordering on $A$ coming from inclusion: $\alpha \leq \alpha ' \Leftrightarrow T_\alpha \subset T_{\alpha '}$. Choose a maximal totally ordered subset $A' \subset A$, and let $T' = \bigcup _{\alpha \in A'} T_\alpha $. We claim that $T'$ is irreducible. Namely, suppose that $T' = Z_1 \cup Z_2$ is a union of two closed subsets of $T'$. For each $\alpha \in A'$ we have either $T_\alpha \subset Z_1$ or $T_\alpha \subset Z_2$, by irreducibility of $T_\alpha $. Suppose that for some $\alpha _0 \in A'$ we have $T_{\alpha _0} \not\subset Z_1$ (say, if not we're done anyway). Then, since $A'$ is totally ordered we see immediately that $T_\alpha \subset Z_2$ for all $\alpha \in A'$. Hence $T' = Z_2$. This proves (3). Part (4) is an immediate consequence of (3) as a singleton space is irreducible.
$\square$

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