Consider the following forcing:
$P=Fn(\aleph_0,\aleph_0,\aleph_0)=${$p:$dom$(p)$ $ \to \aleph_0 |$ dom$(p)$ $ \subset \aleph_0,$ card(dom$(p)$) $<\aleph_0$}
Can someone give me a hint of how to prove that this way defined $P$ does not satisfy "forcing axiom", i.e. the following does not hold:
For any set $\mathbb D$ of cardinality $\leq 2^{\aleph_0}$, there exists a filter $H$ on $P$ such that $H \cap D \neq \emptyset$ for every $D \in \mathbb D$ which is dense in $P$.
Let $\mathbb{C}$ denote your forcing above (which is also Cohen forcing).
For each $x \in {}^\omega 2$, Let $D_x = \{p \in \mathbb{C} : (\exists n \in \mathrm{dom}(p))(p(n) \neq x(n))\}$. $D_x$ is dense in $\mathbb{C}$.
Let $E_n = \{p \in \mathbb{C} : n \in \mathrm{dom}(p)\}$. $E_n$ is dense.
Let $\mathcal{D} = \{D_x : x \in {}^\omega 2\} \cup \{E_n : n \in \omega\}$.
Note that any filter $F$ for $\mathbb{C}$ meeting all the $E_n$'s give a single real.
Now $\mathcal{D}$ has size $2^{\aleph_0}$. If your "forcing axiom" holds, then there is a $G$ in your universe $V$ which is a filter which meets all dense sets in $\mathcal{D}$. Let $g$ be the real which coming from $G$. Since $G \in V$, $g$ must be a real of $V$. But $G$ meets $D_g$ which implies $g \neq g$. Contradiction.