Must a topological space with the following property be a manifold? Edit: it has been pointed out to me that the original definitions I gave do not quite work. Here is the revised question:
Let $X$ be a topological space and $\mathscr{B}$ be a basis of $X$. For any subset $U$ of $X$, say that $U$ has an upper bound in $\mathscr{B}$ if there exists $B\in \mathscr{B}$ such that $U\subseteq \overline{B}$. Say that $B'\in \mathscr{B}$ is a least upper bound of $U$ if $U\subseteq \overline{B'}$ and for all $B\in \mathscr{B}$, if $U\subseteq \overline{B}$ then $B'\subseteq B$. Say that $\mathscr{B}$ has the least-upper-bound property if every subset of $X$ which has an upper bound in $\mathscr{B}$ has a least upper bound in $\mathscr{B}$.
Say that $X$ is $\mathscr{B}$-dense if, for any pair of elements $x, y \in X$, the set $\{x, y\}$ has a nonempty least upper bound $B\in \mathscr{B}$, such that $x, y \not\in B$.
I believe that $\mathbb{R}^n$ given the standard basis of open balls should satisfy both these properties—unless I have gotten something wrong somewhere, which is eminently possible.
Now, onto the actual question: suppose that $X$ is a second-countable Hausdorff space, for which there exists a basis $\mathscr{B}$ such that $\mathscr{B}$ has the least-upper-bound property and $X$ is $\mathscr{B}$-dense. Is $X$ necessarily a manifold?
Apologies if this question is rather trivial.
 A: Let $X$ be a T1 topological space (every singleton is closed). Let ${\mathcal B}$ be your basis of topology on $X$; for $x\in X$ set ${\mathcal B}_x:= \{B\in {\mathcal B}: x\in B\}$. Since $X$ is T1,
$$
\bigcap_{B\in {\mathcal B}_x} B =\{x\}. 
$$
Note that each singleton $\{x\}\subset X$ has an upper bound (since ${\mathcal B}$ is a basis of the topology of $X$).
Now, suppose that $B_x$ is a least upper bound of a singleton $\{x\}\subset X$. Since each $B\in {\mathcal B}_x$ is an upper bound of $\{x\}$, it follows that $B_x\subset B$ for all $B\in {\mathcal B}_x$. Since $X$ is T1,
$$
B_x\subset \bigcap_{B\in {\mathcal B}_x} B =\{x\}.$$
Since $B_x$ cannot be empty (as you require $\{x\}\subset \overline{B_x}$), it follows that $B_x=\{x\}$.
Since $B_x$ is open, $X$ has discrete topology. Consider now a subset $\{x,y\}\subset X$. Since you did not assume that  $x\ne y$, we can as well take $x=y$. As I noted earlier, the unique least upper bound $B_x$ of $\{x\}$ is $\{x\}$. But then $x\in B_x$, contradicting your 2nd condition.
Thus, we proved that each T1 space satisfying your axioms is empty.
