 Solution

Countability of the forcing and uncountability of $Y$ are understood internally in $M$ and $M[G]$, respectively. This matters because the model itself is externally countable. Choose a name $\dot{Y}\in M$ and a ground-model set $A\in M$ containing every ground-model element that can occur in $Y$. Such an $A$ exists: the rank of $\dot{Y}$ bounds the ranks of its values, so a sufficiently high $V_{\theta }^M$ contains $Y\subseteq M$.

For $p\in P$, ground-model axiom schema of separation and the forcing definability lemma giveBy the forcing truth lemma,If every $X_p$ for $p\in G$ were countable in $M$, it would remain countable in $M[G]$, where $G\subseteq P$ is countable. Then $Y$ would be countable, contrary to the hypothesis. Thus some $p\in G$ has $X_p$ uncountable in $M$; otherwise its ground-model enumeration would still enumerate it in the extension. Let $X=X_p$. Every one of its members is forced by $p$ into $Y$, soThe countable chain condition for forcing also preserves its uncountability. Separativity is not needed for this particular argument. This is the ground-model uncountable subset lemma for countable forcing.