Product topology is discrete Willard's General Topology says:

For each $\alpha\in A$, let $X_{\alpha}$ be discrete topological space. Then $\prod_{\alpha\in A}X_{\alpha}$(under product topology) will be a discrete space if and only if A is finite.

But, if $X_{\alpha}=\{1\}$ for each $\alpha\in A$, then  $\prod_{\alpha\in A}X_{\alpha}$ is discrte space, right?
 A: The question has been phrased badly.  It should say:

Let $A$ be a set.  Then $A$ is finite if and only if for all collections $(X_\alpha)_{\alpha\in A}$ of discrete topological spaces, the product $\Pi_{\alpha \in A} X_\alpha$ is discrete.  

As your example shows, moving the 'for all $(X_\alpha)$' quantifier outside gives us a sentence that is untrue.  
For one direction, it should be sufficient to consider the case that $X_\alpha=\{0,1\}$ for all $\alpha$.  
A: You're right that the formulation is not optimal.
I think he meant to say: 

For each $\alpha\in A$, let $X_{\alpha}$ be discrete topological space with more than one point. Then $\prod_{\alpha\in A}X_{\alpha}$(under product topology) will be a discrete space if and only if A is finite.

Or

For each $\alpha\in A$, let $X_{\alpha}$ be discrete topological space. Then $\prod_{\alpha\in A}X_{\alpha}$(under product topology) will be a discrete space if and only if $\{\alpha \in A: |X_\alpha| > 1\}$ is finite.

Or as an escape clause: maybe Willard defines somewhere that a discrete space is by definition one that has at least 2 points, or some such trick. I don't have it at hand now.
In simple terms: he wants to state that "all" infinite products of discrete spaces are no longer discrete any more. But he does need to add the clause that the spaces are not singletons to avoid trivialities.
