Proof of Extreme Value Theorem on a Closed Interval

The interval is nonempty and compact. Since $a\le a$ and $a\le b$, the point $a$ belongs to $[a,b]$, so $[a,b]$ is nonempty. The set $[a,b]$ is the set of $x$ with $a\le x$ and $x\le b$, which is the set to which A Closed Interval is Sequentially Compact in the Real Line applies; by that theorem it is sequentially compact in $(\mathbb{R},d_{\mathbb{R}})$, hence compact in $(\mathbb{R},d_{\mathbb{R}})$ by A Sequentially Compact Subset of a Metric Space is Compact.

Continuity gives semicontinuity. By the final assertion of Semicontinuity Under Negation and Characterization of Continuity, a function that is continuous on a subset of a metric space, as a map into $(\mathbb{R},d_{\mathbb{R}})$, is both upper semicontinuous and lower semicontinuous on that subset. Applied to $f$ on $[a,b]$, this shows that $f$ is upper semicontinuous on $[a,b]$ and lower semicontinuous on $[a,b]$.

Attainment. Applying claim 1 of Semicontinuous Functions Attain Their Extrema on a Compact Set to the nonempty compact set $[a,b]$ and the upper semicontinuous function $f$ gives a point $x_{\max}\in[a,b]$ with $f(x)\le f(x_{\max})$ for every $x\in[a,b]$, and claim 2 applied to $f$, which is lower semicontinuous on $[a,b]$, gives a point $x_{\min}\in[a,b]$ with $f(x_{\min})\le f(x)$ for every $x\in[a,b]$. These are the required points.