In a question I found my characteristics curves for my PDE to be $y=-\cos(x)+C$. When I sketched some of the curves I could see that I would fill the $x,y$.
But the question asked explain why your sketch shows that the method of characteristics leads to a solution $u(x,y)$ at all points in the $x, y$ plane.
The only solution I could come up is that $u(x,y)=f(c)=f(y+\cos(x))$. And since $C$ is arbitrary all points in the $x,y$ plane are satisfied.
I don't think it is worthwhile to try and make up a general method to do this. It is better to just come up with something each time, with the guiding aid of a picture (like you did).
In this case, for example, it is evident that every pair $(x_0, y_0)$ belongs to some characteristic curve, because $$y_0=-\cos(x_0)+\underbrace{\cos(x_0)+y_0}_{=C}.$$ Therefore $(x_0, y_0)$ belongs to the curve associated to $C=\cos(x_0)+y_0$.