$X$ is a real Banach space, $p$ is a sublinear functional on $X$ that is nonnegative and lower semicontinuous (in short, $\liminf p(x_n)\ge p(x)$ whenever $x_n\to x$). Then there exists $M<\infty$ such that $p(x)\le M\|x\|$.
I first defined $\phi(x)=\|x\|+p(x)+p(-x)$, and showed that $\phi$ is indeed a norm on $X$ and also dominates the original norm $\|\cdot\|$. Now if only I can show $(X,\phi)$ is complete, I'm done, since this would imply $\|\cdot\|$ and $\phi$ are equivalent and that in particular $p$ is dominated by $\phi$ and thus by $\|\cdot\|$.
I'm now mostly struggling to prove $p(x)$, or $p(x)+p(-x)$ is continuous at $0$. But I don't know how to use the lower semicontinuity here.
That $p$ is lower semicontinuous means that its "closed unit ball", $B_p = p^{-1}([0,1])$ is closed. Since $p$ is sublinear and everywhere finite, $B_p$ is absorbing. By Baire's theorem, $B_p$ has nonempty interior. Since $B_p$ is convex, $0 \in \overset{\Large\circ}{B}_p$, hence there is an $r > 0$ such that $B_{\lVert\cdot\rVert}(0,r) \subset B_p$. By the sublinearity of $p$, it follows that
$$p(x) \leqslant \frac{1}{r}\lVert x\rVert$$
for all $x$.