Let $X$ be a Banach space. Consider $F\colon X\rightarrow (-\infty; \infty]$ and $F$ is l.s.c. Then $S_{\alpha}=\{x\, \colon \, F(x)\leq \alpha\}$ is closed.
Proof: fix $\alpha \in \mathbb{R}\cup\{\infty\}$. Take a sequance $(x_{n})\subset S_{\alpha}$ that is convergent to some $x\in X$. Since $F$ is l.s.c, we have $$\lim_{n} F(x_{n})\geq\liminf_{n}F(x_{n})\geq F(x).$$ On the other hand, $\alpha\geq F(x_{n})\Rightarrow\alpha\geq \lim_{n}F(x_{n})\Rightarrow\alpha\geq F(x)$. Hence $x\in S_{\alpha}$. Am I correct? I think that the fact that $\alpha$ may by $\infty$ is not a problem. What is more, if $F$ is additionaly convex, then $S_{\alpha}$ is convex and by the Mazur theorem weakly closed, which finally implies that $F$ is weak l.s.c. I guess.
Well you are not sure that $\lim_n F(x_n)$ makes sense but you don't need it. Since, as you observed, $\liminf_{n}F(x_{n})\geq F(x)$ and $\alpha\geq F(x_{n}),\forall n$, you conclude $\alpha\geq\liminf_{n}F(x_{n})\geq F(x)$. I think that $\alpha$ must be in $\mathbb{R}$ because $(-\infty,+\infty)$ is not closed in $(-\infty,+\infty]$. For example if $\alpha=\infty$, $F:\mathbb{R}\rightarrow(-\infty,+\infty]$, $F(x):=x$ is continuous but $S_{\alpha}=\mathbb{R}$ is not closed.