This is another, shorter, proof for the fact that $MA_{\aleph_0}$ always holds.

Let $(P,\leq)$ be a partially ordered set and $\mathcal D$ be a collection of subsets of $P$ . We remember that a filter $G$ on $(P,\leq)$ is $\mathcal D$ -generic if $G \cap D \neq \varnothing$ for all $D \in \mathcal D$ which are dense in $(P,\leq)$ . (In this context ``dense'' means: If $D$ is dense in $(P,\leq)$ , then for every $p \in P$ there's a $d \in D$ such that $d \leq p$ .)

Let $(P,\leq)$ be a partially ordered set and $\mathcal D$ a countable collection of dense subsets of $P$ . Then there exists a $\mathcal D$ -generic filter $G$ on $P$ . Moreover, it could be shown that for every $p \in P$ there's such a $\mathcal D$ -generic filter $G$ with $p \in G$ .

Proof. Let $D_1,\dots, D_n, \dots$ be the dense subsets in $\mathcal D$ . Furthermore let $p_0 = p$ . Now we can choose for every $1 \leq n < \omega$ an element $p_n \in P$ such that $p_n \leq p_{n-1}$ and $p_n \in D_n$ . If we now consider the set $G:=\{ q \in P \mid \exists \; n < \omega { s.t. } p_n \leq q \}$ , then it is easy to check that $G$ is a $\mathcal D$ -generic filter on $P$ and $p \in G$ obviously. This completes the proof.