I'm inclined to make this claim because the functions epigraph is $\{(x,t) : t \ge f(x)\}$. But to be a convex cone, it must be closed under the usual
$\theta_1 (x_1,t_1) + \theta_2 (x_2,t_2)$
for $\sum \theta = 1$ and $\theta_i \in [0,1]$. But that then implies:
$\theta_1 t_1 + \theta_2 t_2 \ge f(\theta_1 x_1 + \theta_2 x_2) $
which is the same as
$f(\theta_1 x_1 + \theta_2 x_2) \le \theta_1 f(x_1) + \theta_2 f(x_2)$
implying $f$ is convex.