Locally Compact Metric Space Properties (i) Let $X$ be locally compact, and $K$ compact. Show that, given $\delta > 0$, there exists finitely many compact closed balls of radius at most $\delta$ that cover $K$.  
Proof: Let $\delta >0$. Since $X$ is locally compact, to each $x \in X$ there exists $\epsilon(x) \leq \delta$ such that $C_{\epsilon(x)} (x) : = \{y \in X \, : \, d(x,y) \leq \epsilon(x)\}$ is closed and compact. Now, consider the covering $$K \subset \bigcup_{x \in K} B_{\epsilon(x)} (x) $$ so that $$K \subset \bigcup_{i=1}^{n} B_{\epsilon_i}(x_i)$$ and therefore,
$$K \subset \bigcup_{i=1}^{n} C_{\epsilon_i}(x_i).$$  
(ii) Deduce that if $\epsilon >0$ is small enough, then the set $S:= \{x \in X : d(x,K) \leq \epsilon\}$ with $$d(x,K) = \inf_{y \in K} d(x,y)$$ is compact.   
Proof: The above argument works for any compact subset of $X$, in particular, $\partial K \subset K$, the boundary, is closed and therefore compact. Therefore, we get finitely many closed compact balls that cover $\partial K$, and given $\epsilon < \delta$, cover $S$. Thus it is enough to show that the set $S':= \{x \in X, x \notin K^\circ : d(x,K) \leq \epsilon \}$ is closed, since $$S' \subset S \subset \bigcup_{i=1}^{n} C_\delta (x_i)$$ with $x_i \in \partial K$, and since $d(x,K) = 0$ for any $x \in K$ anyways and thus we can take $S = S' \cup K$, to show $S$ is compact. Since $K$ is compact and metrics are continuous, there is $y \in \partial K$ such that $d(x,K) = d(x,y)$ for each $x \in X$. Now, $S'$ is clearly closed, since if $S' \ni x_n \to x$, then $d(x,y) \leq d(x,x_n) + \epsilon$, and this implies $d(x,y) \leq \epsilon$ as well. Thus, $S'$ is a closed subset of a compact set and therefore compact and so $S$ is compact as the union of two compact sets.  
Note: ($K^\circ$ is the interior of $K$)
Any issues here??  I've been doing measure theory and funct analysis for a little while, so I wanted to do an exercise to keep up on my analysis I/II type skills.
 A: Part $(i)$ is correct. 
For part $(ii)$, you obviously made the argument much complicated than it should be. Geometrically you might want to consider $\partial K$, but in a general metric space and general $K$, you basically knows nothing about $\partial K$ (Yes, intuitions fails here). Working with $K$ will be more or less the same as working with $\partial K$. 
More seriously, you haven't quite shown that $S'$ (or $S$) is compact. Thye problem is that even if $\epsilon <\delta$,
$$S\subset \bigcup_{i=1}^n C_\delta (x_i)$$
might not be true (Counterexample: let $K$ be a line with length $1$ sit inside $X = \mathbb R^2$. Let $x_1, x_2$ be two endpoints of $K$ and $\delta$ a bit bigger then $1/2$. Then $S$ is not contained in $C_{\delta} (x_1) \cup C_\delta(x_2)$ whenever $\epsilon$ is not small enough, in particular when $\epsilon =1/2$). 
Another serious problem is that even if you have the inclusion, $C_\delta (x_i)$ might not be compact, only that $C_{\epsilon _i}(x_i)$ is compact. 
To remedy this, you have to be more careful in part $(i)$. Instead of $B_{\epsilon(x)} (x)$, you use $B_{\epsilon(x)/2} (x)$ as the open cover. Then as $K$ is compact, there are $x_1, \cdots x_n\in K$ so that 
$$K \subset \bigcup_{i=1}^n C_{\epsilon_i/2} (x_i),$$
where $0<\epsilon_i \le \delta$ and each ball $C_{\epsilon_i} (x_i)$ is compact. Now let $\epsilon >0$ be chosen so that $2 \epsilon <\min\{\epsilon_1, \cdots, \epsilon_n\}$. Thus if $y\in S= \{ y\in X : d(y, K)\le \epsilon\}$, there is $x\in K$ so that $d(x, y)\le \epsilon$. As $x\in K$, there is $x_i$ so that $d(x, x_i) \le \epsilon_i/2$. Hence
$$d(y, x_i)\le d(y, x) + d(x, x_i) \le \epsilon + \epsilon_i/2 <\epsilon_i. $$
Thus $y\in C_{\epsilon_i} (x_i)$ and so 
$$S\subset \bigcup_{i=1}^n C_{\epsilon_i} (x_i).$$
As the right hand side is a compact set and $S$ is a closed subset, $S$ is also a compact set. 
