Since $E_1, \dots, E_N$ are closed in $T$, then the complements $X \setminus E_1, \dots, X \setminus E_N$ are open in $T$. By definition, a finite intersection of open sets is open, so the intersection $\bigcap_{n = 1}^N X \setminus E_n$ is open in $T$. Applying
R220: Difference of set and union equals intersection of differences to this intersection, one has
\begin{equation}
\bigcap_{n = 1}^N X \setminus E_n = X \setminus \bigcup_{n = 1}^N E_n
\end{equation}
Since its complement is open in $T$, the union $\bigcup_{n = 1}^N E_n$ is therefore closed in $T$. $\square$
