Countably closed posets don't add a new uncountable W-free set. (Or $\kappa$-Aronszajn trees remain $\kappa$-Aronszajn.) This is a follow up to $PFA \implies SOCA$. (Or using forcing to prove theorems unrelated to consistency.).
Pertinent definitions can be found there.
(Actually, if you prefer, you can skip all the details on PFA and SOCA and just explain to me how to get the $x_\xi$ from the exercise and hint below and how to get the set $\{x_\xi:\xi<\kappa\}$ to be an element of $M$ and it will probably be very useful to me anyway.)
In the second paragraph to that proof, Kunen says that as $\mathbb{P}$ is countable closed, it preserves the fact that no uncountable subset of $E$ is $W$-free. It says it can be argued in the same way as this exercise:

I have an argument but I doubt its validity a bit (Primarily because I'm unsure if I'm "working in $M$"): 
Suppose $p$ forces that $\mathring{C}$ is an uncountable $W$-free subset of $E$. We find a sequence $p\geq p_0\geq...p_\xi\geq...$ for $\xi<\omega_1$, along with elements $x_\xi\in E$ such that $p_\xi\Vdash x_\xi\in \mathring{C}$ and the $x_\xi$ are all different. To do this fix a generic $G$ with $p\in G$. As $M[G]\models |C|=\aleph_1$ there must be some $x_0\in C$ so get $p_0\leq p$ such that $p\Vdash x_0\in C$. Suppose $\xi<\omega_1$ and that we've defined $p_\alpha$ and $x_\alpha$ for $\alpha<\xi$. As this recursion is taking place in $M$, the set $\{x_\alpha:\alpha<\xi\}$ is in $M$ and $M[G]\models C\setminus \{x_\alpha:\alpha<\xi\}\neq \emptyset$ so get a $x_\xi\in C\setminus \{x_\alpha:\alpha<\xi\}$ and a $p_\xi$ that forces $x_\xi\in \mathring{C}$. We can assume this $p_\xi$ is below all the previously defined $p_\alpha$ as the poset is countably closed. This way $\{x_\xi:\xi<\omega_1\}$ is in $M$ (as it was defined by recursion in $M$) and given $x_\alpha$ and $x_\beta$ WLOG assume $\alpha<\beta$, so $p_\beta$ forces that they're both in $C$ and then $(x_\alpha,x_\beta)\notin W$. Which implies $\{x_\xi:\xi<\omega_1\}\in M$ is $W$-free.
Thank you so much for your time. 
 A: Note that you write "fix a generic $G$," and then never use it; this is an indication that you are in fact reasoning inside $M$! Specifically, do away with the generic and just focus on the (definable!) forcing relation itself. Everything I write below takes place inside $M$:
Suppose $\mathbb{P}$ is countably closed, but could add an uncountable $W$-free subset of $E$. Let $\nu$ be a name, and $p$ a condition that forces "$\nu$ is an uncountable $W$-free subset of $E$". Let $\lhd$ be some well-ordering of $\mathbb{P}$, and $\prec$ some well-ordering of $E$ (these exists since $M\models AC$ - for any set in $M$, there's a well-ordering of that set also in $M$). Then we define an $\omega_1$-sequence of conditions and reals as follows:

$p_\alpha$ is the $\lhd$-least condition stronger than $p$ and every $p_\beta$ for $\beta<\alpha$ such that for some $e\in E$, we have $p_\alpha\Vdash e\in \nu$ but $p_\beta\not\Vdash e\in\nu$ for any $\beta<\alpha$; and we let $e_\alpha$ be the $\prec$-least such $e$.

It's a good exercise to show that $p_\alpha$ and $e_\alpha$ are in fact defined for every $\alpha<\omega_1$. You'll use the countable closure of $\mathbb{P}$, together with the fact that each individual condition can only force countably many things into $\nu$ (since otherwise we'd get an uncountable $W$-free set right there).
Then $\{e_\alpha: \alpha<\omega_1\}$ is an uncountable $W$-free subset of $E$, in the ground model (since the above definition by transfinite recursion took place entirely in the ground model).

Note that I used that $M$ satisfies AC, above. What if $M$ doesn't satisfy AC - if I force over a model in which choice fails, do things still work out nicely?
The answer, sadly (or excitingly!), is no. Indeed, we can even have forcings that are countably closed but add reals! For a very silly example of this, let $M$ be a model of ZF which contains an infinite, Dedekind-finite set $D$ (that is, an infinite set into which $\omega$ does not embed) and let $\mathbb{P}_{stupid}$ be the partial order consisting of pairs $p=(\sigma, \tau)$, where $\sigma$ is a finite string of natural numbers, $\tau$ is a finite string of distinct elements of $D$, and $\vert\sigma\vert=\vert\tau\vert$ (order $\mathbb{P}_{stupid}$ in the obvious way).
Then clearly forcing with $\mathbb{P}_{stupid}$ adds a real - forget the $D$ nonsense, the left coordinates are Cohen forcing! But because of the $D$ nonsense, $M$ thinks that every sequence $p_0\ge p_1\ge p_2\ge . . .$ is eventually constant, and hence that $\mathbb{P}_{stupid}$ is countably closed! (See also this question.)
