about a criterion for compacted and connected sets in $\mathbb{R^{n}}$ The next exercise had been difficult for me. Any help is welcome, thanks.
Be $K$ a subset sequentially compact of $\mathbb{R^{n}}$. Proof that $K$ is not connected iff exist two nonempty subsets  $A$ and $B$ of $K$, disjoint, with $A \cup B = K$ and  $\varepsilon > 0$ such that $||u-v||> \varepsilon$ for all $u \in A$ and $v \in B$.
At the end also have to respond this question: is needed the assumption of sequential compactness for the existence of such an $\varepsilon$?
i attempted to prove it first supposing that if this set is not connected then there exists two open sets U and V that separate the set K and supposing that there is no such an $\varepsilon$ that arrives two sequences one in $U \cap K$ and the other in $V \cap K$ that converge to the same point. From there i know that this limit is in $K$ but i don't know where comes the contradiction i need.
 A: The fact is that $A$ and $B$ are closed by the assumption that $K$ is not connected. But closed in compact set implies compact and two disjoint compact set in a metric space have always positive distance. We can use the following:
Lemma: if $A \subset \mathbb{R}^n$ is compact, $B \subset \mathbb{R}^n$ is closed and $A \cap B = \emptyset$ then $d(A,B)>0$.
Proof: the function $f(x)=d(x,B)$ for $x \in \mathbb{R}^n$ is well defined and since A is compact $f$ has a minimum in $A$, which is strictly positive since $A$ and $B$ are disjoint sets.
Remark 1: the "if" proof is clearer. In fact you might consider $U = \{x \in K : d(x,A) < \frac{\varepsilon}{2}$ and $V= U = \{x \in K: d(x,B) < \frac{\varepsilon}{2}$ which are an open partition of $K$
Remark 2: in a metric space sequentially compact and compact are equivalent.
Remark 3: you need compactness because you might think of $A= \mathbb{R} \times \{0\} \subset \mathbb{R} \times \mathbb{R}$ and $B = graph(exp) \subset \mathbb{R} \times \mathbb{R}$. $K = A \cap B$ have not that property.
