Step 1: Simplify the square root.
Notice $x^2-2x+1=(x-1)^2$, so $\sqrt{x^2-2x+1}=|x-1|$.
Step 2: Restate the inequality.
The inequality becomes $|x-1|>x+2$. We handle the absolute value with two cases.
Step 3: Case where x is at least 1.
Here $|x-1|=x-1$, so we need $x-1>x+2$, i.e. $-1>2$, which is impossible. No solutions come from this case.
Step 4: Case where x is less than 1.
Here $|x-1|=-(x-1)=1-x$, so we need $1-x>x+2$.
Step 5: Solve that inequality.
$1-x>x+2\Rightarrow -2x>1\Rightarrow x<-\dfrac{1}{2}$. This is already inside $x<1$, so it stands.
Step 6: Combine the cases.
The full solution set is $\left(-\infty,-\dfrac{1}{2}\right)$, which is option (3).
\[ \boxed{\left(-\infty,-\tfrac{1}{2}\right)} \]