Show that lower semi-continuous function attains it's minimum. (Proof verification) (By contradiction) Let $f: [0,1]\to \mathbb{R}$ be a lower semi-continuous function, then
$$ \liminf_{x\to a} f(x) \geq f(a), \forall a \in [0,1]$$
I have to prove that $f$ attains its minimum on $[0,1]$, that is:
$\exists x_0 \in [0,1]$ such that $f(x_0) \le f(x)$, $\forall x \in [0,1]$.
My proof:
Assume that $f$ is lower semi-continuous, but $f$ doesn't attain it's minimum on $[0,1]$,
Since $f$ is lower semicontinuous at $x_0$,
$$\forall \epsilon > 0, \exists \delta > 0 \mbox{ such that } |x - x_0| < \delta, \Rightarrow f(x) > f(x_0) - \epsilon, \forall x \in (x_0 - \delta,x_0 + \delta)$$
Now since $f$ doesn't attain it's minimum on $[0,1]$, then
$$\forall x_0 \in [0,1], \exists x \in [0,1] \mbox{ s.t } f(x_0) > f(x)$$
Let $\epsilon = f(x_0) - f(x) > 0, \exists \delta > 0$, such that $|x-x_0| < \delta$ and $f(x) > f(x_0) - \epsilon \Rightarrow f(x_0) - f(x) > \epsilon = f(x_0) - f(x)$ which is a contradiction.
I am right? Thanks.

Since $f$ is lower-semicontinuous, for each $x\in [0,1],$ there is an open interval $I_x\subseteq [0,1]$ such that $\inf\{f(y):y\in I_x\}\ge f(x)-1.$ The $I_x$ form an open cover of $[0,1]$ so passing to a finite subcover, we conclude that $f$ is bounded below.
So, letting $y=\inf\{f(x):x\in [0,1]\}$, we can find a sequence $(x_n)\subseteq [0,1]$ such that $f(x_n)\to y.$ And of course,  there is a subsequence $(x_{n_k})\subseteq (x_n)$ such that $x_{n_k}\to x_0\in [0,1]$. 
Then, $f(x_0)\le \liminf_{x\to x_0} f(x)\le \liminf_{k\to \infty}f(x_{n_k})=\lim_{k\to \infty}f(x_{n_k})=y,$ which implies that $f(x_0)=y.$
