I was trying to argue by contradiction. Say there is a $x_0 \in [0,1]$ so that $f(x_0) \neq x_0$. WLOG, assume $f(x_0) < x_0$. Now, notice that $f \circ f$ is the inverse of $f$ by definition. so $f$ has to be monotonic. (increasing or decreasing).
Assume it is decreasing. That is, if $a<b \implies f(a) > f(b)$.
Now, we know there is some $\alpha \in [0,1]$ so that $f(\alpha) = \alpha $. No WLOG assume $\alpha > x_0$, then
$$ f(\alpha) < f(x_0) \implies \alpha < f(x_0) < x_0$$
This is a contradiction.
Is this a valid argument?