Let $T = (X, \mathcal{T})$ be a D1106: Topological space such that

(i)	$F_j \subseteq X$ is a D98: Closed set in $T$ for each $j \in J$
(i)	$\bigcap_{j \in J} F_j$ is the D76: Set intersection of $F = \{ F_j \}_{j \in J}$

To show that the intersection $\bigcap_{j \in J} F_j$ is closed in $T$, by definition, we must show that the complement $X \setminus \bigcap_{j \in J} F_j$ is open in $T$. Applying result R219: Difference of set and intersection equals union of differences, one has $$\begin{equation} X \setminus \bigcap_{j \in J} F_j = \bigcup_{j \in J} (X \setminus F_j) \end{equation}$$ Since $F_j$ is closed in $T$ for each $j \in J$, then $X \setminus F_j$ is open in $T$ for each $j \in J$. By definition of a D86: Topology, an arbitrary union of open sets is open. Thus, $\bigcup_{j \in J} (X \setminus F_j)$ is open in $T$ and therefore $X \setminus \bigcap_{j \in J} F_j$ is open in $T$. By definition of a closed set, then, the intersection $\bigcap_{j \in J} F_j$ is closed in $T$. $\square$