Let $\mu$ denote the joint distribution of $(X,Y,Z)$, and we'll denote the marginal distributions of $\mu$ using subscripts (e.g. $\mu_X$ is the marginal distribution of $X$). Let $\lambda$ denote the Lebesgue measure on $[0,1]^3$.
Typically we think of the joint density of a random variable $(X,Y,Z)$ to be the derivative of the cdf. However, there is another perspective we can look at it. $f$ be the pdf of $\mu$, and Let $g:[0,1]^3\to \mathbb{R}$ be a bounded, measurable function. Then,
$$\mathbb{E}[g(X,Y,Z)] = \int_{[0,1]^3} g(x,y,z) \mu(dx,dy,dz) = \int_{[0,1]^3} g(x,y,z)f(x,y,z)\lambda(dx,dy,dz).$$
Since this is true for all bounded, measurable $g$, $f$ can be represented as the Radon-Nikodym derivative of $\mu$ with respect to $\lambda$:
$$f(x,y,z) = \frac{d\mu}{d\lambda}(x,y,z).$$
Thus, $f$ is only well-defined when $\mu \ll \lambda$. Now consider the following event:
$$\mathcal{E} := \{X\leq 1/2, Y\in [1/2,1], Z = Y\}.$$
Then notice that,
$$\lambda(\mathcal{E}) \leq \lambda(Z=Y) = 0.$$
However,
$$\mu(\mathcal{E}) = \mu(X\in [0,1/2],Y \in [1/2,1]) = 1/4 \neq 0.$$
Thus, $\mu$ is not absolutely continuous with respect to $\lambda$, so $(X,Y,Z)$ does not have a density.
Note: When I'm talking about a density here, I mean in the standard sense with respect to the Lebesgue measure. You could define a representative measure with respect to which $\mu$ is absolutely continuous and use the Radon-Nikodym derivative to get a density with respect to that measure.
Note: An intuitive reason why $(X,Y,Z)$ does not have a density is because it has only 2 degrees of freedom ($X$ and $Y$ determine $Z$) while a density would be used to describe joint distributions with 3 degrees of freedom.
Best Answer
Let's first do it geometrically and then see if that leads to an algebraic insight.
Let $X$ and $U$ be plotted on a coordinate plane, with the $X$-value horizontal and the $U$-value vertical. Then $(X,U)$ is a point in the unit square. If $U \le 1/2$, then $\max(U, 1/2) = 1/2$, and we compare $1/2$ against $X$, meaning that in the lower half of the unit square, the set of points $(X,U)$ satisfying the criterion $\max(U, 1/2) \le X$ comprises a square with vertices $(1/2, 0), (1, 0), (1, 1/2), (1/2, 1/2)$, since any $X \in [1/2, 1]$ will be at least as large as $\max(U, 1/2) = 1/2$.
What happens in the upper half of the square? In this case, $\max(U, 1/2) = U$, and we require $X$ to be at least as large as $U$, so this includes all the points on or below the line $U = X$ in this upper half; i.e., it is the triangle with vertices $(1/2, 1/2), (1, 1/2), (1,1)$.
Combined together, these two regions become a trapezoid which is easily seen to have area $3/8$.
How do we use this to reason algebraically? We simply write
$$\begin{align} \Pr[\max(U, 1/2) \le X] &= \Pr[(1/2 \le X) \cap (U \le 1/2)] + \Pr[(U \le X) \cap (U > 1/2)] \\ &= \Pr[1/2 \le X]\Pr[U \le 1/2] + \Pr[1/2 < U \le X] \\ &= (1/2)(1/2) + (1/8) \\ &= 3/8. \end{align}$$
If $\Pr[1/2 < U < X]$ is not obvious, we can always integrate: $$\Pr[1/2 < U < X] = \int_{u=1/2}^1 \int_{x=u}^1 \, dx \, du = \int_{u=1/2}^1 1-u \, du = \left[u - \frac{u^2}{2}\right]_{u=1/2}^1 = \frac{1}{2} - \frac{3}{8} = \frac{1}{8}.$$