Is every set countable according to some outer model? Let $M$ denote a set-sized model of $\mathrm{ZFC},$ not necessarily well-founded. Is it true that for any $x \in M,$ there is an outer model $O$ of $M$ such that $(O \models x \mbox{ is countable})$?
 A: (Following the clarification in the comments,) I am interpreting an outer model of $M$ as a model $M'$ with the same ordinals that is an end-extension of M. (End-extension: If $M′$ is the outer model, $y\in M$, and $M′\models x\in y$, then $x\in M$.) In that case, the answer is negative in general, if by "there is" one means "in $V$". 
For instance: We could have an $\alpha$ such that $V_\alpha$ is a model of $\mathsf{ZFC}$, and a set $x$ that is uncountable there (and therefore in $V$). If $M'$ thinks that there is a bijection $f$ between $\omega$ and $x$, this gives us a bijection (in $V$) between $\omega$ and $x$, simply setting $\hat f=\{(n,t)\mid M'\models f(n)=t\}$.
Of course, the answer could be yes in some cases: It could be that $M$ is countable, in which case there are (in $V$) models that are (isomorphic to) generic extensions of $M$ where the set in question is countable. Or it could even be that $M$ is uncountable, but "thin". For instance, if $0^\sharp$ exists, then there are many sets that are uncountable in $L$ but countable in $V$. 
If one means instead whether there is such a potential outer model (in a Boolean valued extension of the universe, for instance), then yes: One can simply pass to a forcing extension $V'$ of $V$ where (the transitive closure of) $M$ is countable, and then in $V'$ we have models $M'$ that are (isomorphic to) generic extensions of $M$ where the relevant sets are now countable.
Let me add a short remark, since the result seems interesting. 

Suppose that $M$ is a transitive model of $\mathsf{ZFC}$. Then for every $x\in M$ there is an outer model $M'$ where $x$ is countable iff $\mathsf{ORD}\cap M\le\omega_1$. 

(In the statement above, we also allow the possibility that $M$ is a proper class.)
In one direction, suppose that $\mathsf{ORD}\cap M>\omega_1$. We then have that  $\omega_1^V\in M$ and there is no outer model where $\omega_1^V$ is seen countable. This is as in the second paragraph above. 
Suppose now that $\mathsf{ORD}\cap M\le \omega_1$. Since choice holds in $M$, for every $\alpha<\mathsf{ORD}\cap M$ we have that $V_\alpha^M$ is in bijection (in $M$, and therefore in $V$) with some ordinal of $M$, so each $V_\alpha^M$ is countable, and so is $\mathcal P(V_\alpha^M)^M$. It follows that generics for any poset $\mathbb P\in M$ exist in $V$ (because $\mathbb P\in V_\alpha^M$ for some $M$, and the number of dense subsets of $\mathbb P$ that belong to $M$ is countable in $V$). In particular, given any $x\in M$, this is true for $\mathbb P=\mathrm{Col}(\omega,x)$, the poset that adds a surjection from $\omega$ onto $x$. It follows that if $G\in V$ is $\mathbb P$-generic over $M$, then $M[G]\models x$ is countable (and $M[G]$ is an outer model of $M$).
The fact that choice holds in $M$ is essential here. As explained here, it is consistent to have transitive models $M$ of $\mathsf{ZF}$ of countable height but uncountable size. This implies that for some $\alpha<\mathsf{ORD}\cap M$, we have that $V_\alpha^M$ is uncountable (in $V$ and thus also in $M$), and therefore there is no outer model where it would be seen to be countable.
