en.wikipedia.org/wiki/Martin's_axiom

Let (P) be the poset of nonempty open subsets of ([0,1]) ordered by inclusion, exactly as the paragraph sets it up.

Consider the family
[
F={U\subseteq [0,1]: U\text{ is open and }(0,1/n)\subseteq U\text{ for some }n\in\mathbb N}.
]
This is a filter on (P): it is upward closed, and if (U,V\in F), then for some (m,n) we have ((0,1/m)\subseteq U) and ((0,1/n)\subseteq V), so ((0,1/\max{m,n})\subseteq U\cap V), giving a common lower bound in (F).

Now check the two dense families named in the article:

• For each (x\in[0,1]), the dense set is (D_x={S:x\notin S}).
  If (x=0), every ((0,1/n)) avoids (0). If (x>0), choose (n>1/x); then (x\notin(0,1/n)). So (F\cap D_x\neq\varnothing) for every (x).
• For each (r\in(0,1]), the dense set is (E_r={S:\operatorname{diam}(S)<r}).
  Choose (n>1/r). Then ((0,1/n)\in F) and (\operatorname{diam}(0,1/n)=1/n<r). So (F\cap E_r\neq\varnothing).

Therefore this filter does meet both (\mathfrak c)-sized families. So the article's sentence claiming that any such filter would have to 'simultaneously avoid all points' is incorrect: meeting each dense set only requires the filter to contain some member of each family, not one set that avoids every point at once.

In other words, the theorem (\mathrm{MA}(2^{\aleph_0})) is indeed false, but this particular argument as written does not establish it.

This compactness claim is false as stated.

Using the same poset of nonempty open subsets of ([0,1]) ordered by inclusion, let
[
F={U\subseteq [0,1]: U\text{ is open and }(0,1/n)\subseteq U\text{ for some }n\in\mathbb N}.
]
As noted above, this is a filter. It contains sets of arbitrarily small diameter, because each ((0,1/n)\in F) and (\operatorname{diam}(0,1/n)=1/n\to 0).

But
[
\bigcap F=\varnothing.
]
Indeed:

• (0\notin U) for every (U\in F), since every basic generator ((0,1/n)) omits (0), and every member of (F) is open and contains one of those intervals.
• If (x>0), choose (n>1/x). Then (x\notin(0,1/n)), and ((0,1/n)\in F). So (x) is not in the intersection of all members of (F).

Thus no point lies in every member of (F).

What compactness does give is a nonempty intersection for centered families of closed sets (or for a decreasing sequence of nonempty closed sets), not for arbitrary families of open sets. Replacing the open sets by their closures would fit the standard compactness principle; the sentence in the article skips that essential distinction.