Classical solution to PDE using M.o.C

Consider the transport equation
$$u_y+u^2u_x = 0 tag{1}$$
with initial condition
$$begin{cases}1 &, xleq 0\ 0 &,xgeq 1\sqrt{1-x} &, 0<x<1end{cases} tag{2}$$

Find the solution using the method of characteristics. Up to which time is the solution defined in a classical sense?

So we get

$$begin{align}x_t &= hat{u}^2 \ y_t &= 1 \ hat{u}_t &= 0end{align} tag{3}$$

with

$$begin{align}x_0(s) &= s \ y_0(s) &= 0 \ hat{u}_0(s) &= u(s,0)end{align} tag{4}$$

so we get

$$begin{align}x(t,s)&=hat{u}^2cdot t + s \ y(t,s)& = t \ hat{u}(t,s) &= u(s,0)end{align} tag{5}$$

We now invert the map $(s,t) mapsto (x,y)$:

$underline{sleq0}:$
$x=y+s Rightarrow s=x-y text{with} xleq y tag{6}$

$underline{sgeq 1:}$
$ x=s text{with} xgeq 1 tag{7}$

$underline{0<s<1:}$
$x=(1-s)cdot y + s = y-sy+s = y + s(1-y) quad Rightarrow quad s=frac{x-y}{1-y} tag{8}$
with
$y<x<1 text{if} y<1 text{and} y>x>1 text{if} y>1 tag{9}$

The characteristic are intersecting at the point $(x,y)=(1,1)$. Thus, the solution is defined up to $y=1$. We get:

$$u(x,y)=begin{cases} 1 &, xleq y \ 0 &, xgeq 1 \ sqrt{frac{1-x}{1-y}} &, y<x<1 tag{10}end{cases} quad text{for} quad yleq 1$$

Question: How exactly can I see that the lines intersect at $(x,y)=(1,1)$?