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I am trying to prove the following:

$K \subset U \subset \mathbb R^n$, where $K$ is compact and $U$ open, then there is $\epsilon > 0$, such that $\forall u \in \mathbb R^n \setminus U, \forall k \in K, d(k,u) \ge \epsilon$, where $d$ is the usual metric on $\mathbb R^n$.

I was trying to prove it by contradiction. Assume that $\forall n, \exists k_n \in K, u_n \in \mathbb R^n \setminus U: d(k_n,u_n) < \frac{1}{n}$, then clearly $k_n-u_n \to 0$.

$K$ is compact and thereby constrained, thus $(k_n)_{n \in \mathbb N}$ has a convergent sub-sequence $(k_{n_m})_{m \in \mathbb N}$ which converges in $K$ because $K$ is closed, thus $ (k_{n_m})_{m \in \mathbb N} \to k \in K$.

It means that $u_{n_m} = u_{n_m} - k_{n_m} + k_{n_m} \to k \in \mathbb R^n \setminus U $ , because $\mathbb R^n \setminus U$ is also closed.

But then $k \in K \subset U$ and $k \in \mathbb R^n \setminus U$, so we have got a contradiction.

Have I proved it, or there is an error somewhere?

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    The proof seems fine to me. – 2017-01-30

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Your proof is fine as it is. You can also use the definition of compactness directly: if $x \in K$, then find $r(x)$ such that $B(x,r(x)) \subseteq U$ which can be done as $x \in U$ and $U$ is open. Finitely many $B(x_1,\frac{r(x_1)}, B(x_2, \frac{r(x_2)}{2}),\ldots B(x_N, \frac{r(x_N)}{2})$ also cover $K$ by compactness of $K$. But then $\varepsilon = \frac{1}{2}\min(r(x_1), \ldots r(x_N))$ is as required.

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    Suppose $d(y,x) < \varepsilon$ and $d(x , x_i) < \varepsilon$ then $d(y, x_i) < 2\varepsilon$ so $y \in U$. So you have to half the radii. – 2017-01-30
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    @SergeyZykov $x$ is in $K$, so $x$ is in some $B(x_i, \frac{r(x_i)}{2})$. And if we then assume $d(x,y) < \varepsilon$, then $d(x,y) < \frac{r(x_i)}{2}$ and then $y \in B(x_i, r(x_i)) \subseteq U$, contradiction. – 2017-01-30
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    you write "finitely many $B(x_1,\frac{r(x_1)}, B(x_2, \frac{r(x_2)}{2}),\ldots B(x_N, \frac{r(x_N)}{2})$ also cover $K$". You meant "finitely many $B(x_1,r(x_1)), B(x_2, r(x_2)),\ldots B(x_N, r(x_N))$", did not you? You can not make elements of the initial cover smaller at will. – 2017-01-31