If you add a single Cohen real $X\subseteq\omega$, is there any $Y\subseteq X$ which isn't $X\cap A$ for any $A\in V$? After adding a single Cohen real $X\in V[G]$ through $\text{Add}(\omega,1)$, you have already "accidentally" added $2^{\aleph_0}$-many reals, since any infinite subset of $X$ is necessarily not in $V$ (otherwise, $p\Vdash\check Y\subseteq\dot X$ and so $p(y)=1$ for any $y\in Y$ even though $p$ is finite).
However, I was unable to construct any of these infinite subsets of $X$ without simply using a finite $Z\subseteq X$ and constructing $X\setminus Z$.
Using specification, you can, e.g. take all of the even elements of $X$, however this is inconclusive because there is no way, as I understand it, to guarantee that $X$ has infinitely many even numbers. This goes more generally for any assertion about $X$'s elements that I can think of.
This made me think: the set of all even numbers is an infinite subset of $\omega$ that's already in $V$. If we take another infinite subset of $\omega$ that's already in $V$, say $A$, then $X\cap A$ might hold some promise; you can use sets which don't have anything to do with $X$ to still reduce its size. However, there's no guarantee that this method will produce any infinite subset of $X$. In fact, I'm not even sure to how guarantee that it will produce any.
This leads me to the question: is there necessarily a set $Y\subset X$ which is not $X\cap A$ for any $A\in V$?
 A: Yes, lots of such $Y$s exist.
For example, let $Y$ consist of every other element of $X$; that is, if we write $x_n$ for the $n$th smallest element of $X$, then $Y=\{x_{2n}:n\in\mathbb{N}\}$. Then $Y$ is indeed an infinite subset of $X$, but we don't have $Y=X\cap A$ for any $A\in V$. (This is a simple argument from genericity: break into cases depending on whether $A$ has an infinite complement or not.)
More generally, let $X=\{x_0<x_1<x_2<...\}$. Then for each bi-infinite $E\in \mathcal{P}(\mathbb{N})^V$ let $$X_E=\{x_e:e\in E\}$$ be "$X$'s version of $E$." Then $X_E$ is an infinite subset of $X$, but we won't have $X_E=X\cap A$ for any $A\in V$. (Above, $E$ was the set of even numbers.)
A: Sure.  For instance, consider $Y=\{n\in\omega:n,n+1\in X\}$.  By genericity of $X$, $Y$ intersects every infinite subset of $\omega$ in $V$.  So if $Y=X\cap A$ for some $A\in V$, then $A$ must contain all but finitely many elements of $\omega$.  But also by genericity, $X\setminus Y$ is infinite, so this is impossible.
