The preimage of $(-\infty,a]$ under $f$ is closed for $a \in \mathbb{R}$, then $f$ is semi-continuous. So I've been thinking about this for the last two hours, but I am stuck. 
Suppose $f:X \to \mathbb{R}$ where $X$ is a topological space.
$f$ is said to be semicontinuous if for any $x \in X$ and $\epsilon > 0$, there is a neighborhood of $x$ such that $f(x) - f(x') < \epsilon$ for all $x'$ in the neighborhood of $x$.
The question is the if $f^{-1}((-\infty,a])$ is closed for $a \in \mathbb{R}$, then $f$ is lower semi-continuous. 
I started with choosing an $x \in f^{-1}((-\infty,a])$ and letting $\epsilon > 0$. So far I don't know much characterization of closed sets in a topological space except it is the complements of open sets. 
Not sure if this is correct, but I approached this problem with the idea of nets. Since $f^{-1}(-\infty,a]$ is closed, then for each $x \in f^{-1}(-\infty,a]$, there's a net $\{x_i\}_{i \in I}$ such that it converges to $x$ (not sure if I'm allowed to do that). Pick any neighborhood of $x$ denote by $N_x$ (which will contain terms from $f^{-1}(-\infty,a]$), which contains an open set which has $x$. Let $f(x) = b$. Then $N_{x'}:=[N_x \backslash f^{-1}(-\infty, b)] \backslash [\mathrm{boundary \ of \ this \ set \ to \ the \ left}]$. So this gives me an open set such that it contains $x$ and such that $f(x) - f(x') < \epsilon$ for all $x' \in N_{x'}$. So $f$ must be semi-continuous.
Not sure if there is more to know about closed sets in a topological space, except its complement is open.
Any hint on how to think about this is appreciated.

Here I showed that $f$ is lower semi-continuous whenever $f^{-1}((\alpha,\infty))$ is open for all real $\alpha$. Here we have that $f^{-1}((-\infty,\alpha])$ is closed, so $f^{-1}((\alpha,\infty))=f^{-1}((-\infty,\alpha]^c)=(f^{-1}((-\infty,\alpha]))^c$ is the complement of a closed set, and thus is open.
