Proof of Complements, Unions and Intersections of Closed Sets in a Topological Space

Throughout, a subset $S\subseteq X$ is closed in $X$ exactly when $X\setminus S\in\mathcal{T}$, and we refer to the three numbered conditions in the definition of a topological space.

Claim 1. We have $X\setminus X=\varnothing$ and $X\setminus\varnothing=X$. By condition 1, both $\varnothing$ and $X$ belong to $\mathcal{T}$. Hence $X\setminus X\in\mathcal{T}$ and $X\setminus\varnothing\in\mathcal{T}$, so $X$ and $\varnothing$ are closed in $X$.

Claim 2. Let $n\in\mathbb{N}$ and let $C_1,\dots,C_n\subseteq X$ be closed in $X$. Let $V$ denote the set of all $x\in X$ such that $x\in C_i$ for at least one $i\in\{1,\dots,n\}$. A point $x\in X$ fails to lie in $V$ precisely when $x\notin C_i$ for every $i\in\{1,\dots,n\}$, that is, precisely when $x\in X\setminus C_i$ for every such $i$. Therefore

Each set $X\setminus C_i$ belongs to $\mathcal{T}$, because $C_i$ is closed in $X$. By condition 3 the displayed intersection belongs to $\mathcal{T}$, so $V$ is closed in $X$.

Claim 3. Let $I$ be a set and let $(C_a)_{a\in I}$ be a family of subsets of $X$ with $C_a$ closed in $X$ for every $a\in I$. Let $W$ denote the set of all $x\in X$ such that $x\in C_a$ for every $a\in I$. A point $x\in X$ fails to lie in $W$ precisely when there is some $a\in I$ with $x\notin C_a$, that is, with $x\in X\setminus C_a$. Therefore

The assignment $a\mapsto X\setminus C_a$ is a family of subsets of $X$ indexed by $I$, and each of its members belongs to $\mathcal{T}$ because $C_a$ is closed in $X$. By condition 2 the displayed union belongs to $\mathcal{T}$, so $W$ is closed in $X$. In the degenerate case $I=\varnothing$ this argument still applies: then $W=X$ and the displayed union is $\varnothing$, which belongs to $\mathcal{T}$ by condition 1.