a) Yes, the product $X\times_k Y$ of two varieties over an algebraically closed field $k$ is a variety. To prove it, reduce to affine varieties and for those use:
b) If the field $k$ is algebraically closed and if $A$ and $B$ are $k$-algebras without zero divisors, then their tensor product $A\otimes_kB$ also has no zero divisors.
(Whether $A$ or $B$ is finitely generated is irrelevant.)
This is proved in Iitaka's Algebraic Geometry, page 97 (Lemma 1.54)
Edit: And if $k$ is not algebraically closed...
... all is not lost!
Let $k$ be a field and $R$ be a $k$-algebra which is a domain and with fraction field $K$.
If the extension $k\to K$ is separable and if $k$ is algebrically closed in $K$, then for all $k$-algebras $S$ which are domains, the $k$-algebra $R\otimes_k S$ will be a domain.
(Beware that separable means universally reduced. If $k\to K$ is algebraic, separability coincides with the usual notion.)
With luck, one of $A$ or $B$ may play the role of $R$ and $A\otimes_k B$ will be a domain.
As an illustration, any intermediate ring $k\subset R\subset k(T_1,\cdots,T_n)$ (where the $T_i$'s are indeterminates) satisfies the conditions above and will remain a domain when tensorized with a domain.
Bibliography
For the product of algebraic varieties I recommend Chapter 4 of Milne's online notes.
For the tensor product of fields, you might look at Bourbaki's Algebra, Chapter V, §17.
New Edit
At Li's request I'll show that $X\times Y$ is irreducible.
Let $$X\times Y=F_1\cup F_2$$ with $F_i$ closed and consider the sets $X_i=\lbrace x\in X\mid \lbrace x \rbrace\times Y\subset F_i\rbrace$.
Since $$\lbrace x \rbrace\times Y=[(\lbrace x \rbrace\times Y)\cap F_1]\cup [(\lbrace x \rbrace\times Y)\cap F_2]$$ we see, by irreducibility of $\lbrace x \rbrace\times Y$ (isomorphic to $Y$), that each vertical fiber $\lbrace x \rbrace\times Y$ is completely included in (at least) one of the $F_i$'s. In other words, $$X=X_1 \cup X_2.$$ It suffices now to show that the $X_i$'s are closed, because by irreducibility of $X$ we will then have $X_1=X$ (say) and thus $X\times Y=F_1$: this will prove that indeed $X\times Y$ is irreducible.
Closedness of $X_i$ is proved as follows:
Choose a point $y_0\in Y$. The intersection $(X\times \lbrace y_0 \rbrace)\cap F_i$ is closed in $X\times \lbrace y_0 \rbrace$ and is sent to $X_i$ by the isomorphism $X\times \lbrace y_0 \rbrace \stackrel {\cong}{\to}X$. Hence $X_i\subset X$ is closed. qed.
Best Answer
Here is a reference: on page 136 of the book of Görtz-Wedhorn, Proposition 5.51 (ii) says that a $k$-scheme $X$ is geometrically integral over $k$ if and only if for every integral $k$-scheme $Y$, $X\times_kY$ is integral. This doesn't quite give you what you want immediately, in that there is still an argument to be made, but this is the key ingredient. The fact that $X\times_kY$ is separated and of finite type over $k$ is standard, and presumably not the main point of your question. By the aforementioned result, it is integral. Why is it geometrically integral? Remember we are assuming $Y$ as well as $X$ is geometrically integral over $k$. For any extension $K/k$, $(X\times_kY)\times_k\mathrm{Spec}(K)=(X\times_k\mathrm{Spec}(K))\times_k(Y\times_k\mathrm{Spec}(K))$. Since $X$ is geometrically integral, $X\times_k\mathrm{Spec}(K)$ is as well (geometric integrality is a definition which is obviously insensitive to an extension of the base field), and since $Y$ is geometrically integral, $Y\times_k\mathrm{Spec}(K)$ is in particular integral, so, invoking the result again, we get that $(X\times_kY)\times_k\mathrm{Spec}(K)$ is integral. As $K$ was arbitrary, we win: $X\times_kY$ is geometrically integral, and hence a variety in the sense given.