If $E$ is Lebesgue measurable and non-empty, but $A$ is not measurable, then $E\times A$ is not measurable in $\mathbb{R^2}$?

Question

I feel like this isn’t true. I came up with a counterexample but I’m not sure if this is valid since I am only a beginner.

Let $E$ be the singleton $\{0\}$. Then $E$ is measurable. Let $A$ be the Vitali set in $[0,1]$, then $A$ is not measurable. But $E\times A$ = Vitali set on $y$-axis in plane, which is null. Hence, as null sets are measurable, $E\times A$ is measurable. Hence the claim is false.

Is this argument valid?