Let $I$ be a set (possibly infinite), and, for each $\alpha\in I$, let $X_{\alpha}$ be a closed subset of $\textbf{R}$. Show that the intersection $\bigcap_{\alpha\in I}X_{\alpha}$ is also closed.
MY ATTEMPT
We need to show that $X = \bigcap_{\alpha\in I}X_{\alpha}$ contains all of its adherent points.
Let $X = \bigcap_{\alpha\in I}X_{\alpha}$ and $x$ be an adherent-point of $X$.
Therefore there is a sequence of points $(a_{n})_{n=m}^{\infty}$ entirely contained in $X$ which converges to $x$. Since each set $X_{\alpha}$ is closed and $a_{n}\in X_{\alpha}$ for every $n\geq m$ and $\alpha\in I$, this means that $x\in X_{\alpha}$ for every $\alpha$. Thus $x\in X$.
Hence we conclude that $X$ is closed indeed, and we are done.
Could someone tell me if I am reasoning correctly?
You can have another approach:
We know that if $X_{\alpha}$ is closed then $X_{\alpha}^c$ is open in $\mathbb{R}$. Hence the arbitrary union of $X_{\alpha}^c , {\alpha}\in I$ is $\bigcup_{\alpha\in I} X_{\alpha}^c$ which is open. By De Morgan's Law $\bigcup_{\alpha\in I} X_{\alpha}^c =\{\bigcap_{\alpha\in I} X_{\alpha}\}^c$. Therefore its complement is $\bigcap_{\alpha\in I} X_{\alpha}$, which is closed in $\mathbb{R}$.