Show that if $f: [0,1]\to \mathbb{R}$ is lower semi-continuous then attains its minimum on $[0,1]$

Here is a simple solution.

Let $\{x_n\}\in [0,1]$ be a minimizing sequence, that is, such that

$$\lim_{n\to \infty} f(x_n)= f^*= \inf_{x\in [0,1]}f(x).$$ Such a sequence always exists by the definition of infimum. Since $[0,1]$ is compact, the sequence $\{x_n\}$ has a convergent subsequence. Without loss of generality assume that $x_n\to \bar{x}\in [0,1].$ Then

$$f(\bar{x})\leq \liminf_{n\to \infty}f(x_n)= \lim_{n\to \infty} f(x_n)= f^*.$$ Since $\bar{x} \in [0,1]$ and by the definition of $f^*,$ we can only have

$$f(\bar{x})=f^*,$$ as desired.