Adding a real destroys strategic closure The following should have an elementary argument, but I haven’t been able to think of it.
Suppose $P$ is a nontrivial partial order in $V$.  Let $r$ be a real not in $V$.  Show that in $V[r]$, $P$ is not $\sigma$-strategically-closed.
This question is a natural generalization of a question that came up in my research.  I can prove it if $r$ is added by a forcing $Q$ such that $Q^2$ is c.c.c.  The argument uses Unger's Theorem that for such $Q$, if $\dot R$ is a name for a $\sigma$-strategically-closed poset, then $Q*\dot R$ has the $\omega_1$-approximation property.
 A: Edit: The result here implies that $P$ is not $\sigma$-strategically closed in $V[r]$ if $r$ is added by a ccc forcing $Q$ (thus slightly weakening the condition "$Q^2$ is ccc" in your question), see below.

This is just a partial answer, showing that winning strategies witnessing $\sigma$-strategically-closedness cannot be approximated in $V$. To be precise:
If $V\subseteq W$ are models of $\mathrm{ZFC}$ with $\mathbb R^V\subsetneq\mathbb R^W$ then a winning strategy $\tau$ witnessing some $\mathbb P\in V$ to be $\sigma$-stategically closed cannot be approximated in $V$ in the following sense:
There is no function $T\in V$ mapping descending sequences $\vec p=\langle p_i\mid i< 2n+1\rangle$ in $\mathbb P$ of odd length to an antichain below $p_{2n}$ so that $\tau(\vec p)\geq a$ for some element of $T(\vec p)$.
This just means that "$\tau$ cannot be bounded in $V$ by antichains'' (the reason antichains come up is that $\tau$ is trivially bound by $T(\vec p)=\{q\mid q\leq p_{2n}\}$). Intuitively, any such $\tau$, if it exists, must make very large jumps.
So suppose that $T\in V$ were such an approximation. To ease notation, we will assume that $T$ depends only on the last element, so we can understand $T$ as a function on $\mathbb P$ (however slight adjustments of the following argument give the more general statement). Also, we can assume that $T(p)$ is always a maximal antichain below $p$ with at least two elements. Now, in $V$, given some $p\in \mathbb P$ we will label the elements of $T(p)$ with either $l_p$ or $r_p$ in some way so that both labels appear (think of this as a binary left or right decision on how to proceed after hitting $p$).
If $\vec p=\langle p_i\mid i<N\rangle$ ($N\leq\omega$) is a descending sequence through $\mathbb P$ then we decode an $l-r$ sequence $\mathrm{dec}(\vec p)$ from this: Start with $p_0$. If $\vec p$ is eventually below some (unique!) element $a_0$ of $T(p_0)$ then $\mathrm{dec}(\vec p)$ starts with $l$ or $r$ depending on whether $a$ has label $l_{p_0}$ or $r_{p_0}$. Now $\vec p$ may or may not eventually be below some element $a_1$ of $T(a_0)$ and if it is, $\mathrm{dec}(\vec p)$ continues depending on the $a_0$-label of $a_1$, etc. If the sequence reaches length $\omega$ we stop. Now suppose $\mathrm{dec}(\vec p)$ did indeed reach length $\omega$. Then $\vec p$ can only have a lower bound in $\mathbb P$ if $\mathrm{dec}(\vec p)\in V$: $V$ can define $\mathrm{dec}(\vec p)$ from $p_0$ and any lower bound $q$ of $\vec p$ by induction by using $q$ as a guide on whether $\vec p$ went left or right at each step.
We can use this to turn $\tau$ into a winning strategy $\tau^\ast$ for player $II$ in the Banach-Mazur game on $2^\omega$ in $W$ with payoff set $(2^\omega)^W\setminus(2^\omega)^V$: Remember that players alternate playing finite $0-1$ sequence which we interprete as $l-r$-sequences. We will describe $\tau^\ast$ now: Suppose $\delta$ is an opponent and suppose $\delta$ opens with $s_0$. By starting with $p_0=1_\mathbb P$ we can construct a descending sequence
$\vec p^0=\langle p_i\mid i\leq m_0\rangle$ in $\mathbb P$ with

*

*$\forall i<m_0\ p_{i+1}\in T(p_i)$

*$\mathrm{dec}(\vec p^0)=s_0$
Now $\tau$ has a reply $q_0$ to $p_{m_0}$ (considered as a first move by player $I$) and now we can further strengthen it to some $p_{m_0+1}\in T(p_{m_0})$. The reply $s_1$ of $\tau^\ast$ is simply $l$ or $r$ depending on the $p_{m_0}$ label of $p_{m_0+1}$.  We continue doing this: If $s_2$ is the next move of $\delta$ then we lengthen $\langle p_i\mid i\leq m_0+1\rangle$ to a sequence $\vec p^1=\langle p_i\mid i\leq m_1\rangle$ still satisfying 1. and 2. (with all 0's replaced by 1's) and choose our reply simiarly by looking at $q_1=\tau(p_{m_0}{}^\frown q_0{}^\frown p_{m_1})$.
This produces two infinite sequences
$$s=s_0^\frown s_1^\frown s_2^\frown \dots$$
and
$$\vec p^\ast = p_0\geq p_1\geq p_2\geq\dots$$
with $\mathrm{dec}(\vec p^\ast) = s$. Now $\vec p^\ast$ is itself not necessarily according to $\tau$, but from the construction above it is clear that the subsequence $\langle p_{m_i}\mid i<\omega\rangle$ constitute plays of player $I$ in a game that is according to $\tau$. Hence $\vec p^\ast$ has a lower bound, $s\in V$ and $\tau^\ast$ is indeed winning.
The upshot of this is that $\mathbb R^V$ is comeager in $W$, but this can only happen if $\mathbb R^V=\mathbb R^W$, contradiction.

Now let us see that $\mathbb P$ is not $\sigma$-strategically closed in a ccc extension $W$ with new reals. So suppose otherwise and let $\tau$ witness this. By the above, it is enough to show that, in $V$, $\tau$ is bounded by antichains to get a contradiction. Since $\tau$ lives in a ccc extension, $\tau(\vec p)$ can be covered in $V$ by countable subsets of $\mathbb P$ (uniformly in $\vec p)$. Thus it is enough to turn countable subsets $\{p_n\mid n<\omega\}\subseteq\mathbb P$ into antichains $A\subseteq\mathbb P$ so that any $p_n$ is above some $a\in A$.
First we will solve the problem in $W$, where $\mathbb P$ is $\sigma$-strategically closed. It is easy to see that
$$D=\{p\in\mathbb P\mid \forall n\ p_n\not < p\}$$
is dense open (start with some $q$ and strengthen it repeadetly making sure it is not strictly above $p_n$ for one $n$ at a time and follow $\tau$ so that there is a lower bound in the end). Still in $W$ we can build the desired antichain one step at a time inductively: Suppose we are at step $n$ and have constructed a finite antichain $A_n\subseteq D$ so far so that for all $m<n$ there is $a\in A_n$ with $a_i\leq p_m$. Now let $a_n\leq p_n$ be in $D$. If $A_n\cup\{a_n\}$ is an antichain this is our $A_{n+1}$. Otherwise, find some $a\in A_n$ compatible with $a_n$ and replace $a$ in $A_n$ by some $b\leq a, a_n$. We may have to strenghten some conditions infinitely many times in this construction. If we are careful we follow $\tau$ along the way to make sure that there is a lower bound in the end.
Thus we have an antichain $A\in W$ bounding all $p_n$'s. But the existence of such an antichain is absolute between $V$ and $W$: Consider the tree $S$ consisting of finite sequences $\langle a_i\mid i<n\rangle$ so that

*

*All $p_m$, $m<n$ are above some $a_i$, $i<n$.

*$\{a_i\mid i<n\}$ is an antichain

Ordered by endextension, this is indeed a tree of height $\omega$ in $V$ that has a branch in $W$. By absoluteness of wellfoundedness, $S$ has a branch in $V$ as well and the set of conditions appearing in this branch forms the antichain we were looking for in $V$.
