Talk:Continuous Image of Compact Space is Compact/Corollary 3

Since $f[S]$ compact, then $f[S]$ is closed, hence $f[S]=\overline{f[S]}$. Let $U\ni x$ be an open neighborhood of $\alpha\,\colon=\sup f[S]$. Then there is an $r>0$ such that $B_r(\alpha)=(\alpha-r,\alpha+r)\subseteq U$. From Characterizing Property of Supremum of Subset of Real Numbers[1], there is a $y\in f[S]$ such that $\alpha-r0$ such that $B_r(\alpha)=(\alpha-r,\alpha+r)\subseteq\mathbb{R}\setminus f[S]$. By the characterization of the supremum in the real numbers, there is a $y\in f[S]$ such that $\alpha-r<y\leq\alpha$. Hence $y\in(\mathbb{R}\setminus f[S])\cap f[S]=\emptyset$ which is a contradiction.
 * 1) Characterizing Property of Supremum of Subset of Real Numbers []