The goal is to prove that $\mathsf{MA}$ implies the following. Let $\{ f_{\alpha} : \alpha \in A\}$ be some family of functions from $\omega$ to $\omega$ of size $<2^{\aleph_0}$. Then, there is a function $f:\omega \to \omega$ that is eventually above all $f_{\alpha}$, i.e. ther exists an $N < \omega$ such that $\forall n > N$, $f(n) > f_{\alpha}(n)$.
This seems like an application of Booth's Lemma which states
Let $\{A_\alpha \subset \omega: \alpha \in Y\}$ be a family of subsets of $\omega$ with size less than $<2^{\aleph_0}$. If, for each finite subset $F \subset Y$, we have $\bigcap_{\alpha \in F} A_{\alpha}$ being infinite, then there exists a subset $A \subset \omega$ such that $A \setminus A_{\alpha}$ is finite for all $\alpha \in Y$.
I'm aware that $\mathsf{MA}$ implies Booth's Lemma, and my intuition seems to be the fact that $A \setminus A_{\alpha}$ being finite as this seems to be the desired result if we can translate $A$ to the function $f$, $A_{\alpha}$ as the functions $f_{\alpha}$, and the fact that $A \setminus A_{\alpha}$ is finite to $f$ is less than $f_{\alpha}$ on only finitely many values. One issue that I'm coming across when trying to set this up is the fact that $\bigcap_{\alpha \in F} A_{\alpha}$ must be infinite.
Any hints on how to get started?
One of the key points in using forcing axioms is identifying a suitable forcing notion, and a suitable collection of dense open sets.
Here, the forcing notion is easy: Hechler reals. Namely, a forcing condition is an ordered pair, $(p,F)$ where $p$ is a finite function $p\colon n\to\omega$ for some $n<\omega$, and $F$ is a finite family of functions $\omega\to\omega$. The order is defined by $(q,F')\leq(p,F)$ if $p\subseteq q$ and for any $f\in F$, and any $n$ in the domain of $q\setminus p$, $q(n)>f(n)$.
It is easy to see that this forcing notion satisfies the countable chain condition: if $F$ and $F'$ are two finite families of functions, then $(p,F)$ and $(p,F')$ are compatible. In other words, incompatibility can only be achieved by changing the stem/working part/Cohen part; and there are only countably many of those.
And so, we are left to find suitable dense open sets, but that's fairly immediate: just take $D_f$ to be all the conditions which extend $(\varnothing,\{f\})$.
Finally, apply Martin's Axiom.