Continuous Image of Compact Space is Compact/Corollary 3/Proof 2

Proof
By Continuous Image of Compact Space is Compact, $f \sqbrk S$ is compact.

From Compact Metric Space is Complete and Compact Metric Space is Totally Bounded, $f \sqbrk S$ is complete and totally bounded.

A Totally Bounded Metric Space is Bounded.

Hence both the supremum and the infimum of $f \sqbrk S$ exist in $\R$.

Because $f \sqbrk S$ is complete:
 * $\sup f \sqbrk S \in f \sqbrk S$

and:
 * $\inf f \sqbrk S \in f \sqbrk S$