Real solution of the equation $(x^2-2x+2)^2-2(x^2-2x+2)+2 = x$ 
Calculate all real solutions $x\in\mathbb{R}$ of the equation
  $$
\tag1(x^2-2x+2)^2-2(x^2-2x+2)+2 = x
$$

My Attempt:
I used the concept of a composite function. Let $f(x) = x^2-2x+2$. Then equation $(1)$ converts into $f(f(x)) = x$. Both $f(x) = x$ and $f(x) = -x$ satisfy the given composite function.
Case 1: If $f(x) = x$, then
$$
x^2-2x+2=x\\
x^2-3x+2=0\\
x\in\{1,2\}
$$
Note that $1,2\in\mathbb{R}$.
Case 2: If $f(x)=-x$, then
$$
x^2-2x+2=-x\\
x^2-x+2=0\\
x=\frac{1\pm \sqrt{1-8}}{2}\notin \mathbb{R}
$$
So only $x\in\{1,2\}$ are the real solutions of above equation.
Is my process correct? Is there is any other method by which we can solve the above question?
 A: To the question

Is my process right?

Unfortunately, no, it is not.  There are two problems with your logic:
First, while it is true that the functions $f(x)=x$ and $f(x)=-x$ both (separately) obey the identity $f(f(x))=x$, you appear to be trying to use the converse of this, which is false:  Given that $f(f(x))=x$, it is not necessarily true that either $f(x)=x$ or $f(x)=-x$.  Other possibilities include $f(x)=A-x$ (for any real number $A$) and $f(x)=C/x$ (for any real number $C$, and assuming $x \ne 0$).  Nor is this an exhaustive list of the possibilities; you are essentially trying to describe all involutions (functions that are their own inverse function); a complete characterization of such functions can be found at this question.
But the second problem is perhaps more serious:  You seem to be confusing the statement that $f(f(x))=x$ for a particular value of x with the statement that $f(f(x))=x$ for all x.  Suppose (just for example) that you have a function for which $f(3)=-3$.  What does that tell you about $f(f(3))$?  You know that $f(f(3))=f(-3)$, but you can't conclude that this equals $3$, because we don't know anything about $f(-3)$.
Edited to add:
On the other hand, there is something right about your solution.  If you are fortunate enough to find a value of $x$ for which $f(x)=x$, then it will certainly by true that $f(f(x))=x$ as well, and therefore solving the equation $f(x)=x$ can find some of the solutions to a problem like this.  But it's not guaranteed to find all of them.  The original equation is a 4th degree polynomial, and so in principle one can expect there to be up to 4 real solutions, but this heuristic can only find at most 2 solutions.
In this problem it happens to be the case that this heuristic actually does find all solutions, which seems to me to be a surprising coincidence.  I am not sure off the top of my head how typical this is.  Anybody want to take a shot at classifying those polynomials $f$ for which $f(f(x))=x$ if and only if $f(x)=x$?
A: Setting $T=x-1$ you have $(T^2+1)^2-2(T^2+1)+1-T=0$ or $T^4-T=0$ that is
$T(T-1)(T^2+T+1)=0$ so for real solutions you have $T=0$ or $T=1$, that is $x=1$ or $x=2$.
