Continuous functions on a closed interval attain their extrema

If $a=b$, the singleton range consists only of $f(a)$, so the conclusion holds. Suppose $a<b$. For each natural $n\ge 1$, put $U_n=\{x\in [a,b]:\mathrm{\mid }f(x)\mathrm{\mid }<n\}$. These sets cover $[a,b]$ by [L2]. Each is relatively open: at any $x\in U_n$, the margin $n-\mathrm{\mid }f(x)\mathrm{\mid }>0$ and [L3] gives a neighborhood where $\mathrm{\mid }f(t)-f(x)\mathrm{\mid }$ is smaller than that margin; [L4] then gives $\mathrm{\mid }f(t)\mathrm{\mid }<n$. By [L1] a finite subcollection covers; taking its largest index shows $\mathrm{\mid }f\mathrm{\mid }$ is bounded. Consequently $M=\sup f([a,b])$ exists by [A1].

If $M$ is not attained, for every $n\ge 1$ define $V_n=\{x\in [a,b]:f(x)<M-1\mathrm{/}n\}$. For each $x$, $M-f(x)>0$; by [L2] choose $n$ with $1\mathrm{/}n<M-f(x)$, so the $V_n$ cover. Each $V_n$ is relatively open by [L3] and the positive margin $M-1\mathrm{/}n-f(x)$. A finite subcover and its largest index $N$ imply $f(x)<M-1\mathrm{/}N$ for all $x$, making $M-1\mathrm{/}N$ an upper bound for $f([a,b])$. This contradicts that $M$ is the least upper bound. Thus the maximum is attained.

The function $-f$ is continuous because $\mathrm{\mid }(-f(t))-(-f(x))\mathrm{\mid }=\mathrm{\mid }f(t)-f(x)\mathrm{\mid }$. Applying steps 1.1 and 2.1 to $-f$, its attained maximum is the attained minimum of $f$.