Let $$\mathbb{R}^\omega$$ be the infinite product space $$\mathbb{R} \times \mathbb{R} \times ...$$, and let $$A$$ be the subset of $$\mathbb{R}^\omega$$ containing points with coordinates $$0$$ except for a finite amount of coordinates. The following proof contains an error:

The set $$A$$ can be broken up into subsets of the form $$\prod X_\alpha$$ where $$X_\alpha$$ is $$\mathbb{R}$$ for a finite amount of $$\alpha$$ and $$\{0\}$$ otherwise. Of course, $$A$$ is then the union of all such sets.

Take an arbitrary coordinate $$\beta$$. We can find a subset $$\prod X_\alpha \subset A$$ as described before such that $$X_\beta = \mathbb{R}$$. Since the choice of $$\beta$$ was arbitrary, we can conclude that in the union of all such subsets, each coordinate maps to $$\mathbb{R}$$ and therefore $$A$$ can be expressed as $$\prod X_\alpha$$ where $$X_\alpha = \mathbb{R}$$ for all $$\alpha$$.

Let $$x = (1, 1, ...)$$. Since each $$x_\alpha$$ is in $$\mathbb{R}$$ we can conclude $$x \in A$$ for the derived definition of $$A$$. However, $$x \notin A$$ for the given definition of $$A$$, a contradiction.

I don't see a reason why the definition of $$A$$ cannot exist, so there must be something wrong with the proof, but I can't find the problem.

The issue is with your claim

$$(*)\quad$$ and therefore $$A$$ can be expressed as $$\prod X_\alpha$$ where $$X_\alpha = \mathbb{R}$$ for all $$\alpha$$.

Essentially, all you've shown up until this point is that $$A$$ isn't equal to any of its "finite-length approximations." But this doesn't imply that $$A=\prod_{\alpha\in\omega} \mathbb{R}$$ (and so your final paragraph is a non-starter, since $$x\not\in A$$).

If you try to prove $$(*)$$, I suspect you'll quickly see the issue.

Noah pointed out the mistake. But let me make an even stronger point:

The union of products need not be a product.

To see this, note for example that $$\{\langle x,y\rangle\in\Bbb R^2\mid |x|,|y|<1\}$$ is the union of countably many sets of the form $$(a,b)\times(c,d)$$. But the unit disc itself is not a product of any two subsets of $$\Bbb R$$.

Even worse, $$\Bigr((0,2)\times(0,2)\Bigl)\cup\Bigr((3,5)\times(3,5)\Bigl)$$ is a union of two products and it is not a product, since $$\langle 1,4\rangle$$ is not in the set, and if it were a product of two sets $$A$$ and $$B$$, then $$1\in A$$ since $$\langle 1,1\rangle$$ is in the set, and $$4\in B$$ for a similar reason.