The solutions to the PDE $u_x + u_y = 0$ are functions that are constant in the $(1,1)$ direction.
Recall how you use the initial data $u_0$ to define the solution: for any point $(x,y)$ you follow the characteristic line back to the initial condition at $y=0$ (so the point $(x-y,0)$), and since it was constant along the characteristic, the answer is $u(x,y) = u_0(x-y)$. Since the initial data is defined on $[0,\infty)$, can do this for all points $x\ge y$ i.e. points below or on the line $y=x$.
Now have a look at this graph of the boundary condition, initial condition and characteristic lines-
How would you now define the values of the solution above the $y=x$ line?
(Full solution.) It seems that you don't have any issue using just the "initial condition", i.e. the points $\color{red}{\{(x,0): x\ge 0\}}$. You can also use the "boundary data" i.e. the points $\color{blue}{\{(0,y): y\ge 0\}}$ to define the full "initial curve"
$$ \Gamma = \color{blue}{\{ (0,y, u_{\text{bdry}}(y)) : y\ge 0 \}} \cup \color{red}{\{ (x,0, u_{\text{init}}(x)) : x\ge 0\}} $$
You can parameterise this using $s\in\mathbb R$,
$$ \Gamma(s) =
\begin{cases}
(s,0, \color{red}{u_{\text{init}}(s)}) & s\in [0,\infty) \\
(0,-s, \color{blue}{u_{\text{bdry}}(-s)}) & s\in (-\infty,0]
\end{cases}$$
with a compatability condition $\color{red}{u_{\text{init}}(0)} = \color{blue}{u_{\text{bdry}}(0)}$.
Now, we implement the method of characteristics. To find the characteristic curves $(x,y) = (x(t),y(t))$, we solve
$$ \frac{d}{dt} u(x(t),y(t)) = 0$$
Of course, $ \frac{d}{dt} u(x(t),y(t)) = u_x(x(t)) x'(t) + u_y(y(t)) y'(t) $ so we just need $x'(t) = y'(t) = 1$. Thus all characteristic curves are given by
$$ (x(t),y(t)) = (x_0,y_0) + (t,t) , \quad t\geq t_0 $$
where $t_0$ is the time we leave the set $\Gamma$. Using the same parameter $s$ as before, we determine $t_0$ using $x_0=x_0(s),y_0=y_0(s)$,
\begin{align} s\ge 0 &\implies x_0 = s,y_0 = 0, t_0 = 0 \\
s<0 &\implies x_0 = 0, y_0 = -s, t_0 = -s\end{align}
and the full solution is $$u(s,t) = u(x(s,t),y(s,t)) = \begin{cases} \color{red}{u_{\text{init}}(s)} & s\ge 0, t\ge 0 \\ \color{blue}{u_{\text{bdry}}(-s)} & s < 0, t>-s \end{cases}$$
We have defined the solution on the set $\{ (s,t) : s \in \mathbb R, t \ge \max(-s,0) \} $, and they are related to $x,y$ by
$$x = s+t , y=t $$
So that $s = x-y$,
and it should be clear that this defines the region $x\geq 0$ and $y\ge 0$. In particular $\color{red}{x>y>0}$ iff $\color{red}{s>0}$, and $\color{blue}{y>x>0}$ iff $\color{blue}{s<0}$. Therefore, the answer in the original coordinates is
$$ u(x,y) = \begin{cases} \color{red}{u_{\text{init}}(x-y)} & 0\le y\le x \\ \color{blue}{u_{\text{bdry}}(y-x)} & 0 \le x<y\end{cases}$$
(Final remark.) There are many different ways to write this down. However, the graph above captures the main idea that has to be expressed by all proofs of this fact. Regardless of whichever proof you learn, my opinion is that you should make sure that your understanding matches the graph.
First, we can drop $x_0$ as it doesn't entail the graph of the curves to change, then apply the initial conditions.
$$\quad x(t) = t; \quad y(t) = \frac{1}{2}t^2+y_0; \quad u(t)=t+u_0$$
$$y= \frac{1}{2}x^2+y_0; \quad u=x+u_0$$
Now $u(3,y)=y^2$, so is $3+u_0=\left(\frac{1}{2}3^2+y_0\right)^2$, bringing us the desired relation between $y_0$ and $u_0$... to get rid of them!
$y_0=y-\frac{1}{2}x^2; \quad u_0=\left(\frac{1}{2}3^2+y_0\right)^2-3$
$u=x+\left(\frac{1}{2}3^2-\frac{1}{2}x^2+y\right)^2-3$
Best Answer
Modifying the problem. Rather than consider the PDE $u_{xt}+u\ u_{xx}+\frac{1}{2}u_{x}^2=0$ with initial condition $u(x,0)=u_0(x)$ as asked above, I will consider the following variant. $$ \text{Solve }u_{xy}+u\ u_{xx} + u_x^2=0\text{ subject to }u(x,0)=f(x).\qquad(\star) $$ There are three differences between this question and that which was asked originally.
Only (1) represents a significant modification of the problem. It makes the solution more tractable and enables it to be found using an elementary application of the method of characteristics. For these reasons, it is conceivable that this was the intended question.
Note: I will not delve into regularity of the solutions in this answer.
Reduction to a first order quasilinear PDE. Write the equation as $$ \frac{\partial}{\partial x}\left(u_y+u\ u_x\right)=0. $$ Thus $(\star)$ is equivalent to $$ u_y+u\ u_x=g(y),\qquad u(x,0)=f(x),\qquad (\star\star) $$ where $g(y)$ is an arbitrary function of $y$ (with sufficient regularity).
Method of characteristics. Perhaps the simplest formulation of the method of characteristics is for quasilinear first order PDEs. These are PDEs of the form $$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u).$$ To solve this equation, one regards the solution as a surface $z=u(x,y)$ in $xyz$-space. Let $s$ parametrize the initial curve $\bigl(s,0,f(s)\bigr)$ and let $t$ be a second parameter, which can be thought of as the distance flowed along a characteristic curve emanating from $\bigl(s,0,f(s)\bigr)$.
The characteristic equations are then $$ \frac{dx}{dt}=a(x,y,z),\quad \frac{dy}{dt}=b(x,y,z),\quad \frac{dz}{dt}=c(x,y,z). $$ Returning to our equation $(\star\star)$, this reduces to $a(x,y,u)=u$ and $b(x,y,u)=1$ and $c(x,y,u)=g(y)$. Thus $$ \frac{dx}{dt}=z,\quad \frac{dy}{dt}=1,\quad \frac{dz}{dt}=g(y) $$ with initial conditions $x(0)=s$ and $y(0)=0$ and $z(0)=f(s)$.
The solution to this system is $$ x=s+zt,\quad y=t,\quad z=f(s)+h(t), $$ where $h(t)$ is the antiderivative of $g(t)$ satisfying $h(0)=0$. Since $g$ was arbitrary, so is $h$ given $h(0)=0$.
The solution. Now we eliminate all occurrences of $t$ by replacing them with $y$, then eliminate $s$ by writing $s=x-zy$. Finally, replace $z$ with $u$ to obtain the implicit equation $$ \boxed{u=f(x-uy)+h(y)}, $$ where $h(y)$ is any sufficiently regular function satisfying $h(0)=0$. This is an implicit equation for the general solution of $(\star)$.
TL;DR. Change the $\frac{1}{2}$ in the original question to $1$ to obtain a PDE solvable by the method of characteristics.