I'm trying to use net to prove this well-known result. Could you have a check on my attempt?
Let $E$ be a compact topological space and $f:E \to \mathbb R$ lower semi-continuous. Then $f$ attains the minimum on $E$.
I posted my proof separately as an answer below, so I can accept my own answer and thus remove my question from unanswered list. If other people post answers, I will happily accept theirs.
If we first consider the open cover $O_n = \{x \in E\mid f(x) > n\}, n \in \Bbb Z$, (which is decreasing), we get that there is a finite subcover by compactness, and hence a subcover of size one ($\{O_n\}$ for the minimal $n$ used), which ensures that that $n$ is a lower bound for $f$ on $E$ and so $\alpha = \inf_{x \in E} f(x)$ is well-defined.
Assume that this $\alpha$ is not the desired minimum (so it's not assumed as a value). Then we define a new open cover $U_n =\{x \in E\mid f(x) > \alpha+\frac{1}{n}\}$ of $E$: it's easy to see that this is now indeed an open cover of $E$, which is decreasing in $n$ (so $n > n'$ implies $U_{n'} \subseteq U_n$, and so for a finite subcover, letting $k$ be the largest $n$ used as index then obeys $E=U_k$ (for some fixed $k$) and so $f(x) > \alpha+\frac1k$ on $E$ but this would contradict $m$ being the largest lower bound for $f$. So $\alpha$ must be the minimum and we're done.
I think that's simpler and more direct than your net-based argument.
Let $U_n := \{x \in E \mid f(x) > n\}$. Because $f$ is l.s.c. $U_n^c$ is closed and thus $U_n$ is open. So $(U_n)$ is an open cover of $E$ which is compact. So there is a finite subcover and thus $a := \inf_{x\in E} f(x) \neq -\infty$.
Then there is a sequence $(x_n)$ in $E$ such that $f(x_n) \searrow a$. Because $E$ is compact, there is a subnet $(x_{\varphi(d)})_{d\in D}$ and $x\in E$ such that $x_{\varphi(d)} \to x$. Let $\varepsilon >0$ and $\mathcal N(x)$ be the set of all open neighborhoods of $x$. Because $f$ is l.s.c., we have $$ f(x) \le \liminf _{y \to x} f (y) := \sup _{V \in \mathcal N (x)} \inf _{y \in V \setminus \{x\}} f(y). $$
There is $N$ such that $f(x_n) < a + \varepsilon$ for all $n \ge N$. There is $d_1 \in D$ such that $x_{\varphi(d)} \in V$ for all $d \ge d_1$. There is $d_2 \in D$ such that $\varphi(d_2) \ge N$. There is $d_3 \in D$ such that $d_3 \ge d_2$ and $d_3 \ge d_1$. This implies $x_{\varphi(d_3)} \in V$ and $f(x_{\varphi(d_3)}) < a + \varepsilon$. So $\inf _{y \in V \setminus \{x\}} f(y) < a + \varepsilon$ and thus $$ f(x) \le \sup _{V \in \mathcal N (x)} \inf _{y \in V \setminus \{x\}} f(y) < \sup _{V \in \mathcal N (x)} a + \varepsilon = a + \varepsilon. $$
This inequality holds for any $\varepsilon >0$, so $f(x) \le a$ and thus $f(x)=a$.