Before trying an “automated” method, it's better to have a close look at the problem.
You can first of all observe that if $2x-2<0$, then the inequality obviously holds, because $|x^2-1|\ge0$. So all values $x<1$ will satisfy the inequality.
Now suppose $x\ge 1$. In this case also $x^2\ge1$, so we can write the inequality as
$$
x^2-1\ge 2x-2
$$
which becomes
$$
x^2-2x+1\ge0
$$
or
$$
(x-1)^2\ge0
$$
which is surely true for all $x$, in particular for $x\ge 1$.
Conclusion: the inequality holds for all real $x$.
The "automated" method has been treated in other answers; to summarize it, you can divide the inequality into two and you'll put together the solution sets at the end
$$
\begin{cases}
x^2-1\ge0\\
x^2-1\ge 2x-2
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
x^2-1<0\\
1-x^2\ge 2x-2
\end{cases}
$$
Simple algebraic manipulation transforms them into
$$
\begin{cases}
x\le -1 \text{ or }x\ge1\\
(x-1)^2\ge0
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
-1<x<1\\
x^2+2x-3\le0
\end{cases}
$$
and, going on,
$$
x\le -1\text{ or }x\ge 1
\qquad\text{or}\qquad
\begin{cases}
-1<x<1\\
-3\le x\le 1
\end{cases}
$$
whence the same conclusion as before. (Note: the brace is another way of saying “and”.)