if X is complete, and $f^2$ is a contraction prove that $f $ has a unique fixed point Let $f: X \to X $ such that for $f^2 $ we have the contraction property, and $X$ is a complete metric space. Prove that f has a unique fixed point.
I have already seen a correct proof of what I am asking, but still I haven't been able to analyse the exact answer I gave in an exam, and especially, I am not able to see a mistake, and maybe there are several mistakes. So maybe someone can help me with that?
The "proof" I gave is:
First we see that there exists at least a fixed point $a \in X$:
Let's suppose, by contradiction, that a fixed point doesn't exist, meaning
$f(a) \neq a $ for all $a \in X$ . As $ f(a) - a \neq 0 $, then 
$ d(f(f(a)), f(a)) \le \alpha d(f(a),a) $.
As $X$ is complete, then $f(X) \subset X $. So,
$f^2(a) \subset f(a) $ for some $a \in X$. 
As $X$ is complete and the contraction property holds, $f$ is continuous. This imples that
$ \lim_{x\to a} f(a) = a $, and
$ \lim_{x\to a} f(f(a)) = f (\lim_{x\to a} f(a)) = f(a) $ .
Then, $ d(f(f(a)), f(a)) \le \alpha d(f(a),a) $ 
$\Rightarrow d(f(a), a) \le \alpha d(f(a),a)$ ,
but $\alpha \in (0,1) \Rightarrow d(f(a), a) > \alpha d(f(a),a) $. This is a contradiction. Then it must be that $f(a) =a$.
Now we see that $a$ is the only fixed point:
Let's suppose that there exists $b \in A$ such that $b \neq a$ and 
$ f(b) = b $, 
$f(a) = a$. $f$ is a contraction, so
$ d(a, b) = d(f(f(a)), f(b)) \le \alpha d(a,b) $ . This is imposible, because if $\alpha \in (0,1) $ then $d(a,b) > \alpha d(a,b)$
because $d(a,b) > 0 $. This contradicts the hypothesis, so it shoud be that if $b $ is a fixed point of $f$, then  $a=b$.
 A: (Mistakes in red, comment in red below the mistake)
First we see that there exists at least a fixed point $a \in X$:
Let's suppose, by contradiction, that a fixed point doesn't exist, meaning
$f(a) \neq a $ for all $a \in X$ . As $\color{red}{ f(a) - a \neq 0 }$, then 
$\color{red}{\textit{makes no sense in a metric space}}$
$\color{red} {d(f(f(a)), f(a)) \le \alpha d(f(a),a) }$.
$\color{red}{\textit{You seem to be assuming that $f$ is contractive}}$
$\color{red}{\text{As $X$ is complete, then $f(X) \subset X $}}$. So,
$\color{red}{\textit{This is because the codomain of $f$ is $X$, completeness has nothing to do with it}}$
$\color{red}{\text{$f^2(a) \subset f(a) $ for some }a \in X}$. 
$\color{red}{\textit{This makes no sense: there is no inclusion between two points}}$
As $X$ is complete and the contraction property holds, $f$ is continuous. This implies that
$\color{red}{ \lim_{x\to a} f(a) = a }$, and
$\color{red}{ \lim_{x\to a} f(f(a)) = f (\lim_{x\to a} f(a)) = f(a) }$ .
$\color{red}{\textit{Makes no sense: the variable cannot appear as the limit result. From here on, I cannot make sense of anything else. }}$
Then, $ d(f(f(a)), f(a)) \le \alpha d(f(a),a) $ 
$\Rightarrow d(f(a), a) \le \alpha d(f(a),a)$ ,
but $\alpha \in (0,1) \Rightarrow d(f(a), a) > \alpha d(f(a),a) $. This is a contradiction. Then it must be that $f(a) =a$.
Now we see that $a$ is the only fixed point:
Let's suppose that there exists $b \in A$ such that $b \neq a$ and 
$ f(b) = b $, 
$f(a) = a$. $f$ is a contraction, so
$ d(a, b) = d(f(f(a)), f(b)) \le \alpha d(a,b) $ . This is imposible, because if $\alpha \in (0,1) $ then $d(a,b) > \alpha d(a,b)$
because $d(a,b) > 0 $. This contradicts the hypothesis, so it shoud be that if $b $ is a fixed point of $f$, then  $a=b$.
