As long as the equal sign is included in one of the cases, it is fine.

For the question, you get an equation $x + |x-1|=1$, and there are two cases: (1) $x > 1$; (2) $x\leq 1$.  In case (1), you get $x=1$, but that is not in the range, so there's no solution.  In case (2), you get an identity $1=1$, which means all $x\leq 1$ are solutions.  However, the term $\sqrt{2x-1}$ in the equation requires that $x \geq 1/2$, therefore the final solution is $1/2 \leq x \leq 1$.