Does $\partial_t u(x,t) - \partial_x u(x,t) = 0$ have a non-trivial solution?

Given the boundary conditions, there is only the trivial solution, $u\equiv c$ if $T\ge 1$. However, for $T<1$, non-trivial solutions are possible. This can be seen by using the method of characteristics:

Here, the characteristics are easily found, they are the lines parametrized by $s\mapsto (-s, s)$, along these lines the solution of your PDE has to be constant. By ignoring the particular choice of $T$ and the boundary condition at $x=0$, we get the unique solution: $$u(x,t) = \begin{cases}u_0(x+t),\quad &\text{for }0<x<1-t, \\ c,\quad &\text{for }0<x\ge 1-t \end{cases}.$$ Now we need to also take the B.C. at $x = 0$ into account, which is fulfilled iff (according to the above representation of $u$) $c = u(0,t) = u_0(t)$ for $t<T$. If $u_0$ doesn't fulfill this condition, there is no solution to the problem at all. If $T\ge 1$, this means that $u_0 \equiv c$ is necessary, and $u\equiv c$ is the only solution, which is trivial. If $T<1$, we are free to choose $u_0(x)\neq c$ for $1>x>T$, without conflicting the left boundary condition.

For your first equation, the characteristics are given by $dx/dt = -1$. So on $t$-vs-$x$ plane, they "travel" backwards with slope 1.

Thus, if there exists $x \in (0,1)$ such that $u_0(x) \neq c$, then there will be a discontinuity at some $t$ and at $x = 0$. As I understand it, these are sometimes called weak solutions (my terminology might be wrong here).

If you strictly want continuous solutions, then there is only the trivial solution.