proof of characterization of connected compact metric spaces.

First we prove the right-hand arrow: if $X$ is connected then the property stated in the Theorem holds. This implication is true in every metric space $X$, without additional conditions.

Let us denote with $A_{\epsilon }$ the set of all points $z\in X$ which can be joined to $x$ with a sequence of points $p_1,\mathrm{\dots },p_n$ with $p_1=x$, $p_n=z$ and . If $z\in A_{\epsilon }$ then also $B_{\epsilon }(z)\subset A_{\epsilon }$ since given $w\in B_{\epsilon }(z)$ we can simply add the point $p_{n+1}=w$ to the sequence $p_1,\mathrm{\dots },p_n$.This immediately shows that $A_{\epsilon }$ is an open subset of $X$. On the other hand we can show that $A_{\epsilon }$ is also closed. In fact suppose that $x_n\in A_{\epsilon }$ and $x_n\to \overline{x}\in X$. Then there exists $k$ such that $\overline{x}\in B_{\epsilon }(x_k)$ and hence $\overline{x}\in A_{\epsilon }$ by the property stated above. Since both $A_{\epsilon }$ and its complementary set are open then,being $X$ connected, we conclude that $A_{\epsilon }$ is either empty or its complementary set is empty. Clearly $x\in A_{\epsilon }$ so we conclude that $A_{\epsilon }=X$. Since this is true for all $\epsilon >0$ the first implication is proven.

Let us prove the reverse implication. Suppose by contradiction that $X$ is not connected. This means that two non-empty open sets $A,B$ exist such that$A\cup B=X$ and $A\cap B=\mathrm{\varnothing }$. Since $A$ is the complementary set of $B$ and vice-versa, we know that $A$ and $B$ are closed too. Being $X$ compact we conclude that both $A$ and $B$ are compact sets.We now claim that

$$\delta :=\underset{a\in A,b\in B}{inf}d(a,b)>0.$$

Suppose by contradiction that $\delta =0$.In this case by definition of infimum, there exist two sequences$a_k\in A$ and $b_k\in B$ such that $d(a_k,b_k)\to 0$. Since $A$ and $B$ are compact, up to a subsequence we may and shall suppose that $a_k\to a\in A$ and $b_k\to b\in B$. By the continuity of the distance function we conclude that $d(a,b)=0$ i.e. $a=b$ which is in contradiction with the condition $A\cap B=\mathrm{\varnothing }$. So the claim is proven.

As a consequence, given it is not possible to join a point of $A$ with a point of $B$. In fact in the sequence $p_1,\mathrm{\dots },p_n$ there should exists two consecutive points $p_i$ and $p_{i+1}$ with $p_i\in A$ and $p_{i+1}\in B$. By the previous observation we would conclude that $d(p_i,p_{i+1})\ge \delta >\epsilon$.