Is every c.c.c. non-atomic partial order of size $\omega_1$ a union of countable complete suborders? We say that $\mathbb{P}$ is a complete suborder of $\mathbb{Q}$, if it is a suborder, and maximal antichains in $\mathbb{P}$ remain maximal antichains in $\mathbb{Q}$
As the title says, is every c.c.c. non-atomic partial order of size $\omega_1$ a union of countable complete suborders? 
If the answer is no, what are some common c.c.c. forcings which are this way? eg. Suslin trees.
Thanks
 A: In response to Joel's comment, here is an argument showing that it is consistent that every ccc poset of size $\omega_1$ is the union of its nonatomic countable complete subposets.
Suppose $\text{MA}_{\omega_1}$ holds. It follows that there are no Suslin algebras, so any ccc poset must add a real. Furthermore, by a theorem of Pawlikowski, any poset of size $<\mathbf{add}(\mathcal{M})$ that adds a real must add a Cohen real. Therefore any ccc poset of size $\omega_1$ adds a Cohen real. 
Now let $\mathbb{P}$ be such a poset and take $p\in\mathbb{P}$. Let $p\in A\subseteq\mathbb{P}$ be a (countable) maximal antichain. By the fact above, the cone $\mathbb{P}\upharpoonright p$ adds a Cohen real, so the usual forcing theory gives us a complete subposet $\mathbb{Q}\subseteq\mathbb{P}\upharpoonright p$ which is forcing equivalent to $\mathrm{Add}(\omega,1)$. Now, this equivalence and the fact that $\mathbb{Q}$ is ccc yield that there is a countable dense $\mathbb{Q}'\subseteq\mathbb{Q}$. Taking all of this together, $\mathbb{Q}'\cup\{p\}\cup A$ is a countable complete subposet of $\mathbb{P}$ containing $p$.

Janusz Pawlikowski,
  MR 1825187 Cohen reals from small
  forcings, J. Symbolic Logic
66 (2001), no. 1, 318--324.

A: Noah's affirmative answer is correct with the definition that you've given, but if you want to insist that the suborder is also non-atomic, then the answer can be negative. 
One easy way to see this is to note that if $\mathbb{Q}$ is a complete suborder of $\mathbb{P}$, then every forcing extension by $\mathbb{P}$ admits an intermediate forcing extension by $\mathbb{Q}$. So if $\mathbb{P}$ is the forcing notion arising from a Suslin tree, for example, then it is c.c.c., but it has no nontrivial countable complete suborders, since all such suborders are forcing equivalent to adding a Cohen real and forcing with a Suslin tree adds no reals. 
More generally, one can get a concrete example in ZFC+CH. Let $\mathbb{P}$ be the forcing to add a random real, which is c.c.c., but this forcing adds no Cohen reals, and thus can have no nonatomic complete countable suborders. 
A: Suppose $\mathbb{Q}$ is c.c.c. Then every element $a\in\mathbb{Q}$ is contained in a countable maximal antichain, $A_a\subseteq\mathbb{Q}$. Each $A_a$ is a complete suborder of $\mathbb{Q}$ (the only maximal antichain in $A_a$ is $A_a$ itself, and this is a maximal antichain in $\mathbb{Q}$), and $\mathbb{Q}=\bigcup_{a\in A} A_a$.
