Is true that if $V \cong W$ and $Z \cong Y$ then $V\times Z \cong W \times Y?$ Let $V,W,Z,Y$ finite dimensional vector spaces. Is true that if $V \cong W$ and $Z \cong Y$ then $V\times Z \cong W \times Y?$

My attempt:

Suppose it is true: Then there is $f : V \to W$ isomorphism and $g : Z\to Y$ isomorphism.

Them $\dim V = \dim W$ and $\dim Z = \dim Y.$

So define $$h : V\times Z \to W\times Y,$$
$$h(v,z) = (f(v),g(w)).$$

Note that $\dim V \times Z = \dim W\times Y.$

Even more, suppose that $h(v,z) = (0,0).$ Then $f(v) = 0$ and $g(w) = 0$. So since $f$ and $g$ are injectors $v = 0$ and $w = 0.$

So the claim is proved.
There are two ways I would like to think about this problem, which you have essentially highlighted. One is that the function you define is clearly injective and linear, and hence an isomorphism.  

The other way is that it's really easy to know whether two f.d. vector spaces over the same field are isomorphic; you simply have to check whether they have the same dimension (this view is similar to picking a basis). Then, isomorphism follows from the easy-to-prove fact that $$\dim X\times Y=\dim X + \dim Y $$
Yes, this works. If you take the trouble to show surjectivity this will work in arbitrary dimensions as well.