The product of Hausdorff spaces is Hausdorff

I'm confused how it can be true that the product of an infinite number of Hausdorff spaces $X_\alpha$ can be Hausdorff.

If $\prod_{\alpha \in J} X_\alpha$ is a product space with product topology, the basis elements consists of of products $\prod_{\alpha \in J} U_{\alpha}$ where $U_{\alpha}$ would equal $X_{\alpha}$ for all but finitely many $\alpha$'s. If this is the case and we had two distinct points, $x$ and $y$, in $\prod_{\alpha \in J} X_{\alpha}$ and a basis element, $B_x$ containing $x$ then In looking for a basis element,$B_y$ that contains $y$ but is disjoint from $B_x$ then for every $\alpha$ such that the open set $U_\alpha$ of $B_x$ is equal to $X_\alpha$ the open set $U_\alpha$ for $B_y$ would have to be empty. But then it couldn't possibly contain $y$. (or any other point of $\prod_{\alpha \in J} X_\alpha$ for that matter). How then is it possible that $\prod_{\alpha \in J} X_\alpha$ is Hausdorff in the product topology?

What am I missing?

If $x$ and $y$ are distinct points of $\prod_{\alpha\in J}X_\alpha$, then there is at least one $\alpha_0\in J$ on which they differ, meaning that $x_{\alpha_0}\ne y_{\alpha_0}$. $X_{\alpha_0}$ is Hausdorff, so there are open sets $U_{\alpha_0}$ and $V_{\alpha_0}$ in $X_{\alpha_0}$ such that $x_{\alpha_0}\in U_{\alpha_0}$, $y_{\alpha_0}\in V_{\alpha_0}$, and $U_{\alpha_0}\cap V_{\alpha_0}=\varnothing$. Now let $U_\alpha=V_\alpha=X_\alpha$ for each $\alpha\in J\setminus\{\alpha_0\}$, let $U=\prod_{\alpha\in J}U_\alpha$, and let $V=\prod_{\alpha\in J}V_\alpha$; then $U$ and $V$ are basic open sets in $\prod_{\alpha\in J}X_\alpha$, $x\in U$, $y\in V$, and $U\cap V=\varnothing$. The only statement there that might not be immediately evident is that $U\cap V=\varnothing$; to see this, note that if $z\in U\cap V$, then $z_{\alpha_0}\in U_{\alpha_0}\cap V_{\alpha_0}=\varnothing$, so no such $z$ can exist.

Thus, $\prod_{\alpha\in J}X_\alpha$ is Hausdorff.