Is $T(C) \subseteq R^m$ closed?

Big thanks to @Shalop who had the idea of using plain old cones as a counterexample, I just cleaned up the algebra.

So, by rotating the cone defined by $x^2+y^2=z^2;z\geq0$ by $45$ degrees toward the positive $x$-axis, we get the cone $y^2=2xz$ (this is just an exercise in $2D$ rotations). The projection of this cone to the $xy$ plane is the set $$S=\{(x,y):x>0\}\cup\{(0,0)\}$$ This is simple to see. If $x=0$, then $y^2=0$ and so only the vertical ray $(0,0,z);z\geq0$ get projected onto the $y$-axis onto the point $(0,0)$. If $x>0$, we get that the point $(x,y,y^2/2x)$ is on the cone and gets projected into $(x,y)$. This covers all possibilities, so the projection of the cone is $S$ but $S$ is not closed.