A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value

$f[X]$ is a compact subset of $(\mathbb{R}, d_{\mathbb{R}})$, and it is nonempty because $X$ is.

So $f[X]$ is closed in $\mathbb{R}$ and bounded as a subset of the metric space $(\mathbb{R},d_{\mathbb{R}})$: there are $x_0 \in \mathbb{R}$ and a real $\rho > 0$ with $f[X] \subseteq B(x_0,\rho) = (x_0-\rho,\ x_0+\rho)$.

Hence $x_0 + \rho$ is an upper bound and $x_0 - \rho$ a lower bound of $f[X]$, so $f[X]$ is bounded above and below, and being nonempty it has a supremum $u$ and an infimum $\ell$.

For every real $\varepsilon > 0$ there is $t \in f[X]$ with $u - \varepsilon < t \le u$, so $|t - u| < \varepsilon$ and $t \in B(u,\varepsilon) \cap f[X]$; therefore every ball around $u$ meets $f[X]$ and $u$ lies in the closure of $f[X]$.

Since $f[X]$ is closed, $u \in f[X]$; so $u$ is a member of $f[X]$ bounding it above, that is $u = \max f[X]$, and $u = f(x_{\max})$ for some $x_{\max} \in X$.

The same argument with $\ell$ in place of $u$, using the infimum form of step 4.1, gives $\ell \in f[X]$, so $\ell = \min f[X]$ and $\ell = f(x_{\min})$ for some $x_{\min} \in X$.

For every $x \in X$ the value $f(x)$ lies in $f[X]$, hence $f(x_{\min}) = \ell \le f(x) \le u = f(x_{\max})$, which is the assertion, with $f(x_{\max}) = \sup f[X]$ and $f(x_{\min}) = \inf f[X]$.