Proving $\forall f\in \mathbb R ^{\mathbb R}(f\neq i_{\mathbb R})\to (\exists g\in \mathbb R ^{\mathbb R} (f\circ g \neq g\circ f))$

Question

Prove $\forall f\in \mathbb R ^{\mathbb R}(f\neq i_{\mathbb R})\to (\exists g\in \mathbb R ^{\mathbb R} (f\circ g \neq g\circ f))$

Proof by contradiction:

For all $f(x)\neq x$ and $(\forall g\in \mathbb R ^{\mathbb R} (f\circ g = g\circ f))$

Let $f\in \mathbb R ^{\mathbb R}$ such that $f(x)\neq x$ and since the statement is for all $g$ then we can take: $g(x)=5$.

Compose the two functions:

$f(g(x))=f(5)=m, s.t: m\neq 5 \\ g(f(x))=g(a)=5, s.t: a\neq x$

So we have $5=m$ but $m\neq 5 $. contradiction.

Is the $f$ I used general enough?