How to show $P^1\times P^1$ (as projective variety by Segre embedding) is not isomorphic to $P^2$?

One way to see it is to note that $\mathbb{P}^1\times\mathbb{P}^1$ maps onto $\mathbb{P}^1$ (by projection to its first factor) while $\mathbb{P}^2$ does not. In fact, any map $\mathbb{P}^2\to\mathbb{P}^1$ is constant.

[Edit: Based on Georges's comment, I think I should explain why all maps $\mathbb{P}^2\to\mathbb{P}^1$ are constant. If $f:\mathbb{P}^2\to\mathbb{P}^1$ is any non-constant map, then its image is irreducible and has $\dim>0$, hence, it must be dense in $\mathbb{P}^1$. Now in $\mathbb{P}^1$, we can take two points $a\neq b$, and pull them back along $f$. This gives (for most choices of $a$ and $b$) two closed, dimension 1 subvarieties (or curves) in $\mathbb{P}^2$ that do not intersect. This is not possible by Bezout's theorem. You can generalize this argument to show that all maps $\mathbb{P}^n\to X$ with $X$ any variety of dimension $<n$ are constant.]

The surface $\mathbb{P}^1\times\mathbb{P}^1$ has disjoint algebraic curves lying on it, like $\{a\}\times \mathbb{P}^1$ and $\{b\}\times \mathbb{P}^1$ (where $a\neq b\in \mathbb P^1$).
The surface $\mathbb{P}^2$ does not have any disjoint algebraic curves lying on it: this is a very weak form of Bézout's theorem.

The Picard groups of both varieties are not isomorphic (this is more advanced) :
We have $\operatorname {Pic} (\mathbb P^2)\cong \mathbb Z$ whereas $\operatorname {Pic} (\mathbb P^1\times \mathbb P^1)\cong \mathbb Z^2$ (see Ischebeck's article here for a very general result concerning the Picard group of the product of two varieties )