{{explain|It is not clear how the premises of the subproofs (e.g. "$T$ is disconnected and close packed" directly relate to the premises of the main theorem, e.g.: "$S$ is a connected space..."}} === Necessary Condition === Suppose that $T$ is disconnected and a linear continuum. Then there are non-empty open sets $U, V$ that separate $T$. Let $a \in U$ and $b \in V$. {{WLOG}}, suppose that $a \prec b$. Let $B = \set {p \in S: \closedint a p \subseteq U}$ $B$ contains $a$, so it is non-empty, and it is bounded above by $b$. In fact, if $y$ is any point not in $U$, it must be an upper bound for $B$: Because, if it were not bounded above by $y$, there would be some $p \in B$, such that $y < p$.  But $p \in B$ means by definition that $\closedint a p \subseteq U$.  Now, $a < y < p$, so $y \in \closedint a p$ and so $y \in U$, which was not the case by assumption. So any $y \notin U$ is an upper bound for $B$, and so in particular $b$ is. So $B$ has a supremum $m$, as a bounded above, non-empty set has a supremum. This is part of the definition of linear continuum, namely being Dedekind complete. First suppose that $m \in B$. Then $m \in U$. Since $U$ is open, there exists a $p \in S$ such that $\hointr m p \subseteq U$. Since $S$ is close packed, it has an element $q$ strictly between $m$ and $p$. Then $m \in B$ implies that: :$\closedint a m \subseteq U$ We also have that: :$\closedint m q \subseteq \hointr m p \subseteq U$ so: :$\closedint a q \subseteq U$ Then $q \in B$, contradicting the fact that $m$ is an upper bound for $B$. Thus $m \notin B$. So $\closedint a m \nsubseteq U$ (or else $m \in B$ by definition). So there is some element $w \in S$ that is in $\closedint a m$ and that is not in $U$. Thus $w$ lies in $V$, as these two sets cover $S$.  As already proved above, $w \notin U$ implies it is an upper bound for $B$. But then $m$ is the minimal one upper bound by the definition of supremum, so $w < m$ is impossible. We can only have $w = m$. So $m \notin U$, so by necessity, $m \in V$. Then as $V$ is open, we have some $w' < m$ such that $\hointl {w'} m \subseteq V$. Again, by $S$ being close packed, we have some $w''$ with $w' < w'' < m$.  This $w''$ is in $V$, so not in $U$ So $w''$ is an upper bound for $B$ again, and strictly smaller than $m$. This is a contradiction. This shows that the separation cannot exist. {{qed|lemma}} === Sufficient Condition === Suppose that $T$ is not a linear continuum. Then either $T$ is not close packed or $T$ is not Dedekind complete. Suppose first that $T$ is not close packed. Then there are points $a, b \in S$ such that $a \prec b$ and no point lies strictly between $a$ and $b$. Thus: :$X = a^\preceq \cup b^\succeq$ and the components of this union are disjoint. By Mind the Gap: :$a^\preceq = b^\prec$ :$b^\succeq = a^\succ$ where: :$a^\preceq$ denotes the lower closure of $a$ :$b^\prec$ denotes the strict lower closure of $b$ :$b^\succeq$ denotes the upper closure of $b$ :$a^\succ$ denotes the strict upper closure of $a$. Thus these two sets are open sets that separate $T$. Therefore $T$ is disconnected. Suppose next that $T$ is not Dedekind complete. Then there is a non-empty set $B \subset S$ which is bounded above in $S$ but has no supremum in $S$. Let $U$ be the set of upper bounds of $B$ (non-empty by assumption). {{explain|Minor details}} Since $B$ has no supremum, $U$ is open. Let $A = S \setminus U$ be the set of points that are ''not'' upper bounds for $B$. Let $p \in S \setminus U$. Then there is an element of $s \in B$ such that $p \prec s$. Then: :$p \in s^\prec \subset S \setminus U$ Thus $S \setminus U$ is also open. It is non-empty because it contains all elements of $B$. So $U$ and $S \setminus U$ are open sets separating $S$. So $T$ is disconnected. {{qed}}