Q:

Forcing cardinality of a set

I'm studying Shelah's proof (actually written by Uri Abraham) that adding one generic real implies the existence of a Suslin tree (available in this link, I think that freely for everyone.)
The notion of forcing is the set finite functions from $\omega$ to $\omega$, with stronger being "extending", then we construct a tree on $\omega_1$ by defining some functions by using the functions, and ensuring that the result gives us a Suslin tree.
At one point, the claim is that if $X$ is an uncountable anti-chain in the tree (in the generic extension, of course) then there exists $p$ such that $p\Vdash X$ is an uncountable anti-chain.
Then, it says, we can find $q$ stronger than $p$ for which $Y=\{\alpha | q\Vdash\alpha\in X\}$ is uncountable.
That last statement is unclear to me. I'm sensing that this is something relatively simple like a pigeonhole argument, but I'm uncertain how to deduce it.

A:

Consider a fixed generic extension obtained by a generic where $p$ belongs. We have that $X$ is uncountable. For each $\alpha\in X$ there is $q\in G$ that in $V$ forces $\alpha\in X$. We may assume $q\le p$ by further extending if necessary. This shows that $A_\alpha=\{q\mid q\le p\land q$ forces in $V$ that $\alpha\in X\}$ is nonempty for each $\alpha\in X$. Since Cohen forcing is countable, the same $q$ must be in uncountably many of these $A_\alpha$, and we are done.
There is another way of presenting the argument that avoids talking about forcing extensions: For each $q\le p$ let $A_q$ be the set of $\alpha$ that $q$ forces to be in $X$. If each $A_q$ is countable, then (again, because Cohen forcing is countable) there is an $\alpha$ such that any ordinal forced to be in $X$ by a condition extending $p$ must be strictly below $\alpha$. But then $p$ itself must force $X$ to be contained in $\alpha$, contradicting that p forces $X$ to be uncountable. 

