1) For any $x\in\mathbb{R}$, we know that $x\leq|x|$, that $0\leq|x|$ and that $0\leq x^2$. So $x\leq|x|+x^2+1$ and also $x^2\leq|x|+x^2+1$. This number, $x\leq|x|+x^2+1$, is an example of a number that is larger than both $x$ and $x^2$.
2) If $x,y\in\mathbb{R}$ and $x\neq y$, assume without loss of generality that $x
3) For any $x\in\mathbb{R}$, $x+1>x$. So you can always find a number that is larger than any number you can think of. Which means, if you had a maximum real number, you would always be able to find a larger real number, and so it wouldn't be a maximum.
4) Given any negative real number $x<0$, there is a number between it and 0 (this is the point number 2 we just proved), and this number is smaller than 0, so also negative. If there was a largest negative real number, we would be able to find a number also negative larger than it, so it wouldn't be a maximum.