Is this a proper algebraic proof that $Re(z)\leq\left|z\right|$? Given that $z=(x+iy)\in \mathbb{C}$, I'm supposed to show that $Re(z)\leq\left|z\right|$.
This is my attempt:
Note that $\sqrt{Re(z)} = \sqrt{x} \leq \sqrt{x^2} \leq \sqrt{x^2 + y^2}$. 
Since $\sqrt{x^2} = x$, this shows that $x = Re(z) \leq \sqrt{x^2 + y^2} = \left|z\right|$.
Is this proof sound mathematically? How could I make it better? Is there another, more ingenious way you would go about it?
Updated proof (thanks to @Nameless):
Since $Re(z) = x \leq x^2 \leq x^2 + y^2$, then we see that $\sqrt{x^2}\leq\sqrt{x^2+y^2}$. Since $\sqrt{x^2} = x$, it follows that $x = Re(z) \leq \sqrt{x^2 + y^2} = \left|z\right|$, which is what was to be shown.
Is this version better? Maybe I'm wording it incorrectly?
Thanks.

If $\Re z > |z|$, then $|z|^2 = (\Re z)^2 + (\Im z)^2 \ge (\Re z)^2 > |z|^2$, a contradiction; hence $\Re z \le |z|$.
