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 $A_\varepsilon$ the set of all points $z\in X$ which can be joined to $x$ with a sequence of points $p_1,\ldots,p_n$ with $p_1=x$ , $p_n=z$ and $d(p_i,p_{i+1})<\varepsilon$ . If $z\in A_\varepsilon$ then also $B_\varepsilon(z)\subset A_\varepsilon$ since given $w\in B_\varepsilon(z)$ we can simply add the point $p_{n+1}=w$ to the sequence $p_1,\ldots,p_n$ . This immediately shows that $A_\varepsilon$ is an open subset of $X$ . On the other hand we can show that $A_\varepsilon$ is also closed. In fact suppose that $x_n\in A_\varepsilon$ and $x_n \to \bar x\in X$ . Then there exists $k$ such that $\bar x\in B_\varepsilon(x_k)$ and hence $\bar x\in A_\varepsilon$ by the property stated above. Since both $A_\varepsilon$ and its complementary set are open then, being $X$ connected, we conclude that $A_\varepsilon$ is either empty or its complementary set is empty. Clearly $x\in A_\varepsilon$ so we conclude that $A_\varepsilon=X$ . Since this is true for all $\varepsilon>0$ 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 $A\cap B = \emptyset$ . 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 $$ \delta:= \inf_{a\in A, b\in B} d(a,b) > 0. $$ 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 $A\cap B=\emptyset$ . So the claim is proven.

As a consequence, given $\varepsilon<\delta$ it is not possible to join a point of $A$ with a point of $B$ . In fact in the sequence $p_1,\ldots,p_n$ 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 $d(p_i,p_{i+1})\ge \delta > \varepsilon$ .