PDE - find solution to $u_t+(1-u)u_x=0$ with boundary condition Is my solution/reasoning to the following PDE correct? 

If the solution is ok how to solve the last equation in order to find the second shock line?
$$u_t+(1-u)u_x=0 \quad \quad \quad \quad (1)$$ 
$$u(x,0)=\left\{\begin{matrix}
0 & x<0 \\ 
x & 0 \leq x \leq 1 \\
1 & x>1
\end{matrix}\right. \quad \quad \quad \quad (2)$$
My  attempt 

Using method of characteristics I get
$$\frac{dt}{1}=\frac{dx}{1-u} \quad \quad \quad \quad (3)$$
$$x_0+(1-u(x_0,t) \ )t=x \quad \quad \quad \quad (4)$$
thus
$$u(x,0)=\left\{\begin{matrix}
0 & x<0 & along & x=t+x_0\\ 
x & 0 \leq x \leq 1 & along & x= \frac{t+x_0}{1+t}  \\
1 & x>1 & along & x= x_0
\end{matrix}\right. \quad \quad \quad \quad (5)$$
The sketch of the characteristic lines will contain 4 areas:

1) $x=t+x_0$ for  $x_0 < 0$ , decreasing curves at 45 degrees, 
2) vertical curves for $x=x_0$ for $x_0 > 1$, 
3) lower part of the region $0 \leq x \leq 1$ (for small t values) hyperbolic curves going through this region where $x= \frac{t+x_0}{1+t}$, 
4) area between 1) 2) and 3) a "fan" of curves going through (x=0,t=0)


Finding the u value for 4) using as a guess a function $g(\frac{x}{t})$
$$u(x,t)=g(\frac{x}{t}) \quad \quad \quad \quad (6)$$
$$u_x=\frac{g'}{t}, \ \ u_t=-\frac{xg'}{t^2} \quad \quad \quad \quad (7)$$
substituting (7) into (1)
$$ \frac{g'}{t} (1-u-\frac{x}{t^2}) =0  \quad \quad \quad \quad (8)$$
(8) is equal zero if either $g'=0$ or $(1-u-\frac{x}{t^2})=0$, whereby the first one gives only constant solutions which we are not interested. Therefore
$$u=1-\frac{x}{t^2} \quad \quad \quad \quad (8)$$
As the next step I want to find the discontinuity lines (shock lines) between 
A.   2) and 3) 
B.   3) and 4) 
using the Rankine-Hugonoit jump condition
$$\phi(u)'=1-u \quad \quad \quad \quad (9)$$
$$\phi(u)=u-\frac{1}{2} u^2 \quad \quad \quad \quad (10)$$
$$\frac{dx_s}{dt}=\frac{[\phi]}{[u]}=\frac{u^{+}-\frac{1}{2} u^{{+}^2} - u^{-} +\frac{1}{2} u^{{-}^2} }{u^{+}-u^{-}} \quad \quad \quad \quad (11)$$
for the A case I got
$$\frac{dx_s}{dt}=\frac{1}{2} x \quad \quad \quad \quad (12)$$
thus the shock line starting in $(x=1, t=0)$
$$x=ce^{\frac{1}{2} t} \quad \quad \quad \quad (13)$$
and for B case I got the following equation
$$\frac{dx_s}{dt}=\frac{1}{2} \frac{x}{t^2} -1 \quad \quad \quad \quad (14)$$
which I don't know how to solve.



UPDATE

It's assumed $t>0$
I tried to plot the characteristic lines. 
is that correct?

 A: I would only add some results to the Michal's attempt which is basically correct.
$$u_t+(1-u)u_x=0$$
$$\frac{dt}{1}=\frac{dx}{1-u}=\frac{du}{0}$$
A first characteristic comes from $du=0 \quad\to\quad u=c_1$
A second characteristic comes from $\frac{dt}{1}=\frac{dx}{1-u}=\frac{dx}{1-c_1}\quad\to\quad x-(1-c_1)t=c_2$
The solution expressed on the form of implicit equation is :
$$\Phi\left(u\:,\:x-(1-u)t\right)=0$$
where $\Phi$ is any differentiable function of two variables.
Solving the implicit equation for the first variable gives :
$$u=f\left(x-(1-u)t\right)$$
where $f$ is any differentiable function.
Now, we have to determine which particular function $f$ makes the solution agree with the condition :
$$u(x,0)=\begin{cases}
0 \qquad x<0 \\ 
x \qquad 0 \leq x \leq 1 \\
1 \qquad x>1
\end{cases} \quad\to\quad f(x)=
\begin{cases}
0 \qquad x<0 \\ 
x \qquad 0 \leq x \leq 1 \\
1 \qquad x>1
\end{cases}$$
$$u=\begin{cases}
0 && x-(1-u)t<0 \\ 
x-(1-u)t && 0 \leq x-(1-u)t \leq 1 \\
1 && x-(1-u)t>1
\end{cases}$$
$u=x-(1-u)t\quad\to\quad u=\frac{x-t}{1-t}$
$$u=\begin{cases}
0 && \frac{x-t}{1-t}<0 \\ 
\frac{x-t}{1-t} && 0 \leq \frac{x-t}{1-t} \leq 1 \\
1 && \frac{x-t}{1-t}>1
\end{cases}
$$
The graph below is the summary of the results.

IN ADDITION  :
Figure representing the relationship between $t$ and $x$ ( drawing of $t$ as a function of $x$ ) for various constant values of $u$ :
 
Next figure: drawing of $u$ as a function of $x$, for various constant values of $t$ :

Next figure: drawing of $u$ as a function of $t$, for various constant values of $x$ :

A: I just had a look on your Equation 14. What is the problem there? If you replace your "wrong" x by $x_{s}$ on the right hand side, you have a linear ODE of first order, which you can solve by the usual methods. I guess all the solutions might look like
\begin{equation}
x_{s}(t)=C\exp(-\tfrac{1}{2t}) + 1/2\exp(-\tfrac{1}{2t})\operatorname{Ei}(\tfrac{1}{2t})-t
\end{equation}
for $C\in\mathbb{R}$. Now you just need the "initial" condition for the shock curve...and done.
This is exactly
what wolfram alpha computes.
