A sequence converging to a point in a compact Here is the problem: Let $K$ be a compact subset of $ \mathbb{R}^{m} $ ($m>1$) with empty interior and such that $\mathbb{R}^{m}\setminus K $ has no bounded component. For $n=1,2,...$, we define 
$$K_{n}=\lbrace x\in \mathbb{R^{m}}: distance (x,K)=1/n\rbrace.$$ Prove that for all $x\in K$, there is a sequence $(y_{n})$ in $K_{n}$ converging to $x$, as $n\rightarrow\infty$.
Here is what I have done: 
Fix $x\in K$. If for all $r>0$ the ball of center $x$ and radius $r$, $B(x,r)$, encounters some $K_{n}$, then we are done. If not $B(x,r_{0} )\cap K_{n}=\emptyset$, for all positive integer and some $r_{0} >0$. This implies $B(x,r_{0} )$ is included to the complement of $K_{n}$ for all $n$. Intuitively this is impossible since the complements of the $K_{n}$'s encounter each other. How can finish the argument?
 A: Since $K$ has empty interior, for every $r > 0$ there is an $y \in B(x,r/2) \setminus K$. Let $\delta = \operatorname{dist}(y,K)$, and choose $z \in K$ with $\lVert y-z\rVert = \delta$. For $t \in (0,1]$, let $p(t) = ty + (1-t)z$. Then $$\operatorname{dist}(p(t),K) = \lVert p(t) - z\rVert = t\delta$$ since $B(p(t),t\delta)\subset B(y,\delta)$ by the triangle inequality. Hence $B(x,r) \cap K_n \neq \varnothing$ for all $n > 1/\delta$.
Now choose a sequence $(r_k)$ of radii tending to $0$, and corresponding $\delta_k$ such that $r_{k+1} < \delta_k$ for all $k$. For $n \leqslant 1/\delta_1$, choose $y_n \in K_n$ arbitrary. For $1/\delta_k < n \leqslant 1/\delta_{k+1}$ choose $y_n \in K_n \cap B(x,r_k)$. Then $(y_n)$ is a sequence with $y_n \in K_n$ for every $n$, and $y_n \to x$.
A: We fix $x\in K$. The following is a bit along the spirit in your attempt.
The function ${\rm dist} (z,K)$ is (1-Lipschitz) continuous in $z$.
It follows that also
 $$M(r)= \sup \{ {\rm dist}(z,K): \|z-x\|\leq r \}, \; \; r\geq 0$$
 is  (1-Lipschitz) continuous in $r\geq 0$. 
We have $M(0)=0$ (since $x\in K$) and $M(r)>0$ for every $r>0$ (since $K$ is compact and has empty interior). Since $K$ is bounded, $M(r)$ tends to $+\infty$ as $r\rightarrow \infty$. Clearly $M(r)$ is monotone increasing.
By the intermediate-value theorem, for every $n\geq 1$ there is $r_n$ (not necessarily unique) such that $M(r_n)=\frac1n$.
Since $\bar{B}(x,r_n)$ is compact the sup is attained (a maximum) so we conclude that there is $y_n$ such that
    $$ d(y_n,x) \leq r_n \; , \; \; {\rm dist}(y_n,K)=\frac{1}{n}$$
i.e. $y_n\in K_n$. We claim that $r_n\rightarrow 0$ as $n$ goes to infinity:
For $\delta>0$ we have $M(\delta)>0$ so there is $N=N(\delta)\geq 1$ for which  $M(\delta)>\frac{1}{N}$. But then by monotonicity of $M$ for every $n\geq N$ we must have $r_n<\delta$ which proves the claim.
(the condition on the complement not having bounded components is not necessary)
