The product of quasi-projective varieties is a quasi-projective variety. Let 
$$
\varphi_{n,m}:\mathbb{P}^{n}\times \mathbb{P}^{m}\rightarrow \mathbb{P}^{nm+n+m}, ((p_{0}:\ldots:p_{n})(q_{0}:\ldots:q_{m}))\mapsto (p_{0}q_{0}:p_{0}q_{1}:\ldots:p_{n}q_{m})
$$
be the Segre embedding. I have to prove that if $X\subseteq \mathbb{P}^{n},Y\subseteq\mathbb{P}^{m}$ are quasi-projective varieties, then $\varphi_{n,m}(X\times Y)$ is a quasi-projective variety. Let $W\subseteq \mathbb{P}^{n},Z\subseteq\mathbb{P}^{m}$ be closed sets and $U\subseteq \mathbb{P}^{n},V\subseteq\mathbb{P}^{m}$ be open sets such that 
$$
X=W\cap U,
$$
$$
Y=Z\cap V.
$$
We may assume that $X,Y$ are dense in $W,Z$ respectively. According to this, $X$ (resp. $Y$) is irreducible if and only if $W$ (resp. $Z$) is irreducible.
Let 
$$
\pi_{1}: \mathbb{P}^{n}\times \mathbb{P}^{m}\rightarrow\mathbb{P}^{n}
$$
$$
\pi_{2}:\mathbb{P}^{n}\times \mathbb{P}^{m}\rightarrow \mathbb{P}^{m}
$$
be the projections.
I am able to prove that 
$$
\varphi_{n,m}(X\times Y)=[\varphi_{n,m}(\pi_{1}^{-1}(W))\cap\varphi_{n,m}(\pi_{2}^{-1}(Z))]\cap[\varphi_{n,m}(\pi_{1}^{-1}(U))\cap\varphi_{n,m}(\pi_{2}^{-1}(V))]
$$
Hence, $\varphi_{n,m}(X\times Y)$ is a quasi-projective set, because the first set is closed and the second is open. Now, How do we know that it is irreducible?
 A: Assume otherwise. Then there exist two algebraic sets $A \cup B=X \times Y$. Now, given any fiber $X_i=\pi_1^{-1}(x_i)=\{x_i\}\times Y$ of a point $x_i\in X$, $X_i$ is isomorphic to $Y$, hence irreducible. Write $X_i=(A\cap X_i)\cup (B \cap X_i)$. Then either $X_i \subset A$ or $X_i \subset B$. Let $X_A, X_B$ denote the set of $x_i$ such that $X_i \subset A, B$ respectively. If we can prove that $X_A, X_B$ are closed, then, since $X = X_A \cup X_B$, either $X_A$ or $X_B =X$. But, $X_A=\cap \pi_1^{-1}((X\times \{y_i\})\cap A)$ where the intersection ranges over all $y_i\in Y$. Thus $X_A$ is the intersection of closed sets, and hence closed.
A: The Segre embedding gives a homeomorphism of the underlying topological space onto its image, so your question reduces to: is the product of irreducible spaces irreducible?
The answer is: yes. Suppose $U$ is a non-empty open subset of the product; we need to prove it is dense. Replacing $U$ with a smaller set we may assume it is of the form $U_1 \times U_2$, and then the result is clear.
