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 $T$. $\square$
