Closures and Infinite Products

[joeboo]

( This is from an exercise in Munkres' Topology )

Let $X_\alpha$ be an indexed collection of spaces, and $A_\alpha \subset X_\alpha$ be a collection of subsets.
Under the product topology, show that, as a subset of $X = \prod_\alpha{X_\alpha}$

$$\overline{\prod_\alpha{A_\alpha}} = \prod_\alpha{\overline{A_\alpha}}$$

This part I have no problem with. However, he then asks if this holds under the box topology on $X$

Clearly ( closure being the intersection of all containing closed sets ):

$$\overline{\prod_\alpha{A_\alpha}} \subset \prod_\alpha{\overline{A_\alpha}}$$

The reverse inclusion, on the other hand, has me in a twist.

The following is my attempt at a proof:

Let $x = (x_\alpha) \in \prod_\alpha{\overline{A_\alpha}}$, and $U \subset X$ be a neighborhood of $x$. We can assume $U$ is a basis element for the box topology on $X$
Then $\exists \hspace{3} U_\alpha \subset X_\alpha$ such that $U = \prod_\alpha{U_\alpha}$, where $U_\alpha$ are open.

Then we have:
$$x = (x_\alpha) \in U \longrightarrow x_\alpha \in U_\alpha$$

Because the $U_\alpha$ are neighborhoods of the $x_\alpha$, and $x_\alpha \in \overline{A_\alpha}$ for all $\alpha$, we have:

$$U_\alpha \cap A_\alpha \neq \varnothing$$ for all $\alpha$

so that:

$$\prod_\alpha{U_\alpha} \cap \prod_\alpha{A_\alpha} \neq \varnothing$$

That, and $x \in U$ gives:

$$x \in \overline{\prod_\alpha{A_\alpha}}$$

and therefore:

$$\prod_\alpha{\overline{A_\alpha}} \subset \overline{\prod_\alpha{A_\alpha}}$$

I can't see the flaw in this proof, yet I can't somehow shake the feeling that this inclusion shouldn't hold. I know the box topology can give some funky results ( like the closure of the set of sequences with finitely many non-zero entries, or the product of continuous functions not necessarily being continuous ), so I'm a bit weary of it. That, and the way Munkres states the question gives me the suspicion it's a trick question.
So, is there a flaw with my proof? Or am I just being paranoid ( which I often tend to be )

Thanks in advance for any comments, and also for putting up with my anal-latex exactness.

 

[fresh_42]

This part I have no problem with.

I have no proof in mind, but a plan. Write it down (I would do it with open sets instead), and find out where you used the finiteness property of the product topology. Either you didn't use it, in which case the proof holds for the box topology, too, or you see how to construct a counterexample, namely such that this step of the proof is hurt.