This is not true for convex $f$: Take $$ f(x,y)= \max( e^x + y,0), $$ which is a maximum of convex functions, hence convex. The minimum $0$ is attained at, e.g., $(x,y)=(0,-1)$.
Define $A=\\{(x,0): \ x\in \mathbb R\\}$. Minimizing $f$ on $A$ is equivalent to minimizing $e^x$ for $x\in \mathbb R$, where the infimum is not attained.