Hausdorff gaps and $\mathfrak{p}=\mathfrak{t}$ Recently Malliaris and Shelah proved that $\mathfrak{p}=\mathfrak{t}$
(see: http://math.uchicago.edu/~mem/Malliaris-Shelah-CST.pdf). Their results are far more general, however for the specify context of this question what they showed is that if $G$ is an ultrafilter on $\omega$, then
the ultra product $\omega^\omega/G$ is a linear order with no
$(\kappa,\lambda)$-gaps with $\kappa,\lambda<\mathfrak{t}$.
They can obtain the desired equality combining it with a result of Shelah on the existence of a certain type of small gap
(i.e. with both sequences in the gap of size smaller than $\mathfrak{t}$)
in structures of the form $\omega^\omega/G$
in the case $\mathfrak{p}<\mathfrak{t}$.
Recall that

*

*$\mathfrak{t}$ is the least size of a decreasing chain
in $P(\omega)/Fin$ without a positive lower bound in the boolean algebra
$P(\omega)/Fin$.


*$(f_\alpha:\alpha<\kappa,g_\beta:\beta<\lambda)$ is a $\kappa,\lambda$ pregap in a partial order $(X,<^*)$ if  $f_\alpha<^*f_\beta<^*g_\eta<^*g_\gamma$ for all
$\alpha<\beta<\kappa$ and $\gamma<\eta<\lambda$.


*A pregap $(f_\alpha:\alpha<\kappa,g_\beta:\beta<\lambda)$ on $(X,<^*)$
is a gap if for no $f\in X$ we have that $f_\alpha<^*f<^*g_\gamma$
for all
$\alpha<\kappa$ and $\gamma<\lambda$.


*A gap $(f_\alpha:\alpha<\omega_1,g_\beta:\beta<\omega_1)$ on the partial
order $(\omega^\omega,<^*)$ where $<^*$ is eventual domination is an
Hausdorff gap if for all $\beta<\omega_1$ and $n<\omega$
$\{\alpha<\beta:\forall m>n \: f_\alpha(m)<g_\beta(m)\}$
is finite.
Malliaris and Shelah's result has also the following consequence on Hausdorff gaps of which I was not aware and which I ask if it had a proof prior to their result (may be even without extra set theoretic assumptions).

Assume $\mathfrak{t}>\omega_1$,
$(f_\alpha:\alpha<\omega_1,g_\beta:\beta<\omega_1)$ is an Hausdorff gap.
Let $X\subset P(\omega)$ be the set of $A$ infinite subsets of $\omega$
such that
$(f_\alpha\restriction A:\alpha<\omega_1,g_\beta\restriction   A:\beta<\omega_1)$
is no longer a gap in the partial order
$(\omega^A,<^*)$.
Then $X$ is open dense in the Ellentuck topology.

 A: Peter Nyikos and Jerry Vaughan in this paper prove this result, although they don't quite state it that way, and they work in the partial order $(P(\omega),\subseteq^*)$ rather than $(\omega^\omega,<^*)$.
For this partial order, a gap is defined in nearly exactly the same way: just replace the symbol $<^*$ with the symbol $\subseteq^*$ in the definition you gave. Given a gap $(F_\alpha,G_\alpha: \alpha < \omega_1)$, they say that a set $E$ is beside the gap if $(F_\alpha \cap E,G_\alpha \cap E: \alpha < \omega_1)$ is no longer a gap in $(P(E),\subseteq^*)$. The analogy with the situation in your question is (I hope) obvious. They say that a gap is tight if there is no set $E$ beside the gap. They prove (theorem 1.2 of their paper) that
Theorem: There is a tight $(\omega_1,\omega_1)$-gap if and only if $\mathfrak{t} = \aleph_1$.
We'll show in a minute (after proving a lemma below) that this actually implies your statement regarding the Ellentuck topology.
This shows that your proposition (or, at least, a version of it for a different partial order) does depend on the assumption $\mathfrak{t} > \aleph_1$.
As for the proposition in the form you gave it, I don't know whether it had a proof prior to Malliaris and Shelah's work. But it does have a fairly simple proof that doesn't depend on their work. I'll sketch that out now.
Recall that the Ellentuck topology has a basis of sets of the form
$$[s,A] = \{s \cup B : B \in [A]^\omega\},$$
where $s \in [\omega]^{<\omega}$ and $A \in [\omega]^{\omega}$.
Lemma: Suppose $X$ is a subset of $2^\omega$ such that
$\qquad (1)$ if $A \in X$ and $B \subseteq^* A$, then $B \in X$.
$\qquad (2)$ if $B \in [\omega]^{\omega}$, then there is some infinite $A \in X$ such that $A \subseteq B$.
Then $X$ is open dense in the Ellentuck topology.
(To paraphrase: If you're open dense in the partial order $([\omega]^{\omega},\subseteq^*)$, then you're open dense in the Ellentuck topology.)
Proof:
To see that $X$ is open: Suppose $A \in X$. By $(1)$, $[\emptyset,A] \subseteq X$, so $A$ is in the interior of $X$.
To see that $X$ is dense: Let $[s,A]$ be a basic open neighborhood in the Ellentuck topology. By $(2)$, there is some infinite $B \in X$ with $B \subseteq A$. By $(1)$, $s \cup B \in X$. But $s \cup B \in [s,A]$, so $X \cap [s,A] \neq \emptyset$.
QED
Note: this lemma shows that (in the terminology of Nyikos and Vaughan given above) if $\mathfrak{t} > \aleph_1$ and we have a gap $(F_\alpha,G_\alpha:\alpha<\omega_1)$, then
$$\{E \in [\omega]^{\omega} : E \text{ is beside the gap}\}$$
is open dense in the Ellentuck topology. Condition $(1)$ is obvious. Condition $(2)$ follows from the aforementioned theorem of Nyikos and Vaughan, since otherwise their theorem I quoted would be false for the gap $(F_\alpha \cap B, G_\alpha \cap B: \alpha < \omega_1)$ (which is a gap in the partial order $(P(B),\subseteq^*)$).
Now let's check that the set $X$ mentioned in your question satisfies the conditions of this lemma.
Condition $(1)$ is fairly obvious.
For condition $(2)$, fix $B \in [\omega]^{\omega}$. For each $\alpha < \omega_1$, let
$$C_\alpha = \{(m,n) : m \in B \text{ and } f_\alpha(m) < n < g_\alpha(m)\}.$$
Using the definition of a pregap, it's fairly easy to see that $\langle C_\alpha : \alpha < \omega_1 \rangle$ is a decreasing chain in $P(\omega \times \omega)/fin$. Since $\mathfrak{t} > \aleph_1$, there is some infinite $C \subseteq \omega \times \omega$ such that $C \subseteq^* C_\alpha$ for every $\alpha$. Let
$$A = \{m : (m,n) \in C \text{ for some } n\},$$
$$h = \{(m,n) : m \in A \text{ and } n = \min \{k : (m,k) \in C\}\}.$$
Clearly $h$ is a function from $A$ to $\omega$, and, for every $\alpha$, $f_\alpha \upharpoonright A <^* h <^* g_\alpha \upharpoonright A$. We know that $A$ must be infinite, because $C$ is infinite and, for each $m \in A$, $C \cap \{m\} \times \omega$ is finite.
Notice that this proof doesn't depend on the gap being Hausdorff, so you can eliminate that assumption from the statement of the proposition.
