Closure of Infinite Union may not equal Union of Closures

Theorem
Let $T$ be a topological space.

Let $I$ be an infinite indexing set.

Let $\left \langle {H_i} \right \rangle_{i \mathop \in I}$ be a family of subsets of a set $S$.

Let $\displaystyle H = \bigcup_{i \mathop \in I} H_i$ be the union of $\left \langle {H_i} \right \rangle_{i \mathop \in I}$

Then it is not always the case that:
 * $\displaystyle \bigcup_{i \mathop \in I} \operatorname{cl} \left({H_i}\right) = \operatorname{cl} \left({\bigcup_{i \mathop \in I} H_i}\right)$

Proof
Let $\displaystyle H_n \subseteq \R: H_n = \left[{\frac 1 n \,.\,.\, 1}\right]$ for $n \ge 2$.

Then:
 * $\operatorname{cl}\left({H_n}\right) = H_n$

Also:
 * $\displaystyle \bigcup_{n \mathop \ge 2} \operatorname{cl}\left({H_n}\right) = \bigcup_{n \mathop \ge 2} H_n = \left({0 \,.\,.\, 1}\right]$

However:
 * $\displaystyle \operatorname{cl}\left({\bigcup_{n \mathop \ge 2} H_n}\right) = \left[{0 \,.\,.\, 1}\right]$

So:
 * $\displaystyle \bigcup_{n \mathop \ge 2} \operatorname{cl}\left({H_n}\right) \ne \operatorname{cl}\left({\bigcup_{n \mathop \ge 2} H_n}\right)$