Let $U$ be a random variable uniformly distributed over the interval $[0,1$]. We define $X$ such that
$$X = U-0.5 \quad \textrm{if} \quad U - 0.5 \geq 0$$ $$X = 0 \quad \textrm{if} \quad U - 0.5 \leq 0.$$
The task is to find the CDF $F_{X}$.
We know that $X \in [0, 0.5]$.
Let $0 \leq c \leq 0.5$. Then,
$$\begin{align}F_{X}(c) =& P(X\leq c)\\ =& P(U-0.5 \leq c)\\ =& P(U \leq c + 0.5)\\ =& \int_{?}^{c+0.5}du \end{align}$$
Since $U \geq 0.5$, shouldn't the lower limit in the above integral be $0.5$?
Your process seems to confuse the definition of $X$ to be always $U-0.5$. I think with some careful consideration for the limits of $c$ this could work, but I would recommend avoiding the issue altogether by breaking the probability into segments.
$$F_X(c)=P(X\leq c)\\ =P(X\leq c \land U-0.5>0)+P(X\leq c \land U-0.5\leq0)\\ =P(U-0.5\leq c \land U-0.5>0)+P(0\leq c \land U-0.5\leq0)\\ =P(U\leq c+0.5 \land U>0.5)+P(U\leq0.5)\\ =\int_{0.5}^{c+0.5}\ du + 0.5\\ =[(c+0.5)-(0.5)]+0.5\\ =c+0.5$$
Note that this assumes $c\geq 0$; if not, then the probability that converts to an integral should instead be $0$, as would the probability of $0\leq c$.