The answer is negative as can be seen by putting together these two facts:
There is a bounded convex lower semicontinuous functions defined on a closed and convex subset of $\mathbb R^2$ that is not continuous. One example is here. For a large stock of examples, see this.
Every convex, lower semicontinuous and positively homogeneous function on $\mathbb R^n$ is a support function. Theorem 13.2 in Rockefellar's Convex Analysis.
For then by (1), let $f$ be a non-continuous bounded convex lower semicontinuous function on a closed subset $D$ of the plane. Extend $f$ to all of the plane by setting it to $+\infty$ outside $D$. This is still convex and lower semicontinuous and it's not continuous on its effective domain.
Finally, define $g$ on $\mathbb R^3$ as follows: $$ g(x,y,z) = \begin{cases} zf(x/z,y/z) &\text{if }z>0\\ +\infty & \text{if }z\le 0\text{ and }(x,y)\ne 0\\ 0 &\text{if }(x,y,z)=0. \end{cases} $$ This is convex and lower semicontinuous everywhere and not continuous on the effective domain, and it's positive homogeneous. By (2), $g$ is a support function.