It is well known that $MA_{\aleph_1}$ implies that every ccc forcing of size $< 2^{\aleph_0}$ is $\sigma$-linked (in fact - a countable union of filters).
On the other hand, a separative forcing of size $> 2^{\aleph_0}$ is never $\sigma$-linked.
Consider the following statement:
$(\dagger)$ Every ccc forcing of size $2^{\aleph_0}$ is $\sigma$-linked.
Question: Is $(\dagger)$ consistent with ZFC?
I decided not to add further information (for now), because I fear it might cause unnecessary distractions.
The answer is no. Let $\mathbb{P}$ consist of all non empty finite sets $F$ of convergent sequences of reals $c: \omega \to \mathbb{R}$ satisfying: If $c, d \in F$, then $\lim c \notin \{d(n) : n < \omega\}$. Order $\mathbb{P}$ by reverse inclusion. It is a fun exercise to check that $\mathbb{P}$ is ccc but not $\sigma$-linked.
This example appears in S. Todorcevic, Two examples of Borel posets with the ccc, Proc. of the AMS, Vol. 112, No. 4, 1991.