Show that no periodic function is one to one. I would just like to get some feedback if the following proof outline I have written is correct.
Claim. If $f$ is periodic, then $f$ is not injective. 
Proof. Assume to the contrary that $f$ is periodic and $f$ is injective. Then by definition, $\exists$p $\in \mathbb{R}$\ {$0$}, such that $f(x+p) = f(x)$ for all $x\in$ dom $f$. Let $x_1,x_2 \in$ dom $f$. Since $f$ is injective, $f(x_1) = f(x_2)$ and so by definition it follows that $x_1 = x_2$. Now, 
$f(x_1) = f(x_2)$
$\Longrightarrow f(x_1) = f(x_2+p)$; because $f$ is periodic
$\Longrightarrow x_1 = x_2 + p$
$\Longrightarrow x_1 - x_2 = p$
$\Longrightarrow 0 = p$; by assumption that $f$ is injective 
but by assumption $p \neq 0$ which is a contradiction. 
Hence, no periodic function is injective. $\Box$

Your proof has one major flaw: the claim is false! It is easily fixed, however.

Claim: Suppose $f$ is a function on some non-empty subset of the reals. If $f$ is periodic, then $f$ is non-injective.

At that point, you get the existence of a $p$ (as you mentioned) and an $x$! Then $x,x+p$ are distinct elements of the domain of $f,$ but $f(x)=f(x+p),$ so $f$ is not injective.
