proof of Heine-Cantor theorem

We seek to show that $f:K\\to X$ is continuous with $K$ a compact metric space, then $f$ is uniformly continuous. Recall that for $f:K\\to X$ , uniform continuityis the condition that for any $\\varepsilon>0$ , there exists $\\delta$ such that

d_{K}(x,y)<\\delta\\implies d_{X}(f(x),f(y))<\\epsilon

for all $x,y\\in K$

Suppose $K$ is a compact metric space, $f$ continuous on $K$ . Let $\\epsilon>0$ . For each $k\\in K$ choose $\\delta_{k}$ such that $d(k,x)\\leq\\delta_{k}$ implies $d(f(k),f(x))\\leq\\frac{\\epsilon}{2}$ . Note that the collection of balls $B(k,\\frac{\\delta_{k}}{2})$ covers $K$ , so by compactness there is a finite subcover,say involving $k_{1},\\ldots,k_{n}$ . Take

\\delta=\\min_{i=1,\\ldots,n}\\frac{\\delta_{k_{i}}}{2}

Then, suppose $d(x,y)\\leq\\delta$ . By the choice of $k_{1},\\ldots,k_{n}$ and the triangle inequality, there exists an $i$ such that $d(x,k_{i}),d(y,k_{i})\\leq\\delta_{k_{i}}$ . Hence,

	$\\displaystyle d(f(x),f(y))$	$\\displaystyle\\leq$	$\\displaystyle d(f(x),f(k_{i}))+d(f(y),f(k_{i}))$		(1)
		$\\displaystyle\\leq$	$\\displaystyle\\frac{\\epsilon}{2}+\\frac{\\epsilon}{2}$		(2)

As $x, y$ were arbitrary, we have that $f$ is uniformly continuous.
This proof is similar to one found in Mathematical Principles of Analysis, Rudin.