Can somebody please check if my proof is correct? And can it be done in a shorter way? (Using Bolzano Theorem)
A continuous function on a closed, bounded interval $[a,b$ attains a minimum at some point of $[a,b]$.
Solution:
$f$ is continuous and bounded on $[a,b]$ such that $c,d\in[a,b]\subset \mathbb{R}$.
Let $\alpha =inf\left\{f(x):x\in [a,b]\right\}$.
$\implies \alpha+\frac 1n $ is not the greatest lower bound of $f$. Thus, there exists some mapping $x_n \rightarrow f(x_n) \ $ such that $f(x_n)$ lies in between its infimum $\alpha$ and $\alpha+\frac 1n$.
So, $\alpha \leq f(x_n) \leq \alpha+\frac 1n$.
Since $f$ is bounded on $[a,b]$, sequence $(a_n) =\left\{x_n\right\}_{n=1}^{n= \infty } \ \forall n\in\mathbb{N}$, is also bounded on $[a,b]$. Thus by Bolzano-Weierstrass theorem, $a_n$ contains at least one convergent subsequence in the interval $[a,b]$, say $a_{n_i}$ , such that $\lim_{i\to\infty}{ \ (a_{n_i})}=c$
Because f is continuous, we have: $$\lim_{i\to\infty}{ \ (a_{n_i})}=c\Rightarrow f(\lim_{i\to\infty}{ \ (x_{n_i}))}=f(c) \ for \ c\in [a,b]$$
By applying the squeeze theorem for functions, we note:
$$\alpha \leq f(x_n) \leq \alpha+\frac 1n \Rightarrow \lim_{n\to\infty}{ (\alpha)} = \lim_{n\to\infty}{f(x_n)}=\lim_{n\to\infty}{(\alpha+\frac 1n})=\alpha$$
So, $$\lim_{i\to\infty}{f(x_{n_i})=\lim_{n\to\infty}f(x_n)=\alpha}$$
This implies $\alpha :=inf\left\{f(x):x\in [a,b]\right\}=f(c)$
Thus the function f achieves a minimum value $f(c)$ for $c\in [a,b]$.