The answer is negative as can be seen by putting together these two facts:

 1. There are bounded convex lower semicontinuous functions defined on a closed and convex subset of $\mathbb R^2$ that are not continuous. Example [here](https://math.stackexchange.com/questions/2487705/is-a-convex-and-lower-semicontinuous-function-defined-on-a-closed-and-convex-sub).

 2. 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 by (1), let $f$ be a non-continuous bounded convex lower semicontinuous function on a closed subset $D$ of the plane $K = \{ (x,y,z) \in\mathbb R^3 : z = 1 \}$. Extend $f$ to all of $K$ by setting it to $+\infty$ outside $D$. This is still convex and lower semicontinuous and it's not continuous on its effective domain. 

Finally, extend $f$ to all of $\mathbb R^3$ as follows. Let $f(x,y,z)=+\infty$ if $z\le 0$ and $(x,y,z)\ne 0$. Let $f(0,0,0)=0$. Let $f(x,y,z)=zf(x/z,y/z,1)$ if $z>0$. This is still convex and lower semicontinuous on $\mathbb R^3$ and not continuous on the effective domain, and it's positive homogeneous. By (2), our $f$ is a support function.