The answer to the second question is no as well.
Suppose $\dot\tau$ is a $Q$-name for a strategy for the player INC in the game $G_\kappa(P)$. Let us pretend that COM opens the game instead of INC and note that this is unproblematic. We will define by induction a descending sequence $\langle p_\alpha\mid\alpha<\kappa\rangle$ of conditions in $P$ and $Q$-names $\langle \dot p_\alpha\mid\alpha<\kappa\rangle$ so that
- $\Vdash_Q\check p_\beta\leq \dot p_\alpha\leq\check p_\alpha$ for all $\beta\leq\alpha$
- $\Vdash_Q$"The sequence $\langle \dot p_\beta\mid\beta\leq\alpha\rangle$ make up the fist $\alpha+1$-many moves of player COM in a game of $G_{\check\kappa}(\check P)$ in which INC plays according to $\dot\tau$ and the final response of $\dot \tau$ to this is a condition $\leq\check p_\alpha$."
for all $\alpha<\kappa$. It follows that if $G$ is $Q$-generic then COM wins against $\dot\tau^G$ by playing the conditions $\langle \dot p_\alpha^G\mid\alpha<\kappa\rangle$, so $\dot\tau$ is not a winning strategy for INC.
Let us turn to the construction. Assume $\alpha<\kappa$ and that $\dot p_\beta, p_\beta$ are defined for all $\beta<\alpha$. Let $s_0$ be some lower bound of $\langle p_\beta\mid\beta<\alpha\rangle$ which exists as $P$ is ${<}\kappa$-closed. Build descending sequences of maximal possible length
$$\langle s_\gamma\mid\gamma\leq\xi\rangle,\ \langle t_\gamma\mid\gamma<\xi\rangle$$
and an antichain $A_\xi:=\{q_\gamma\mid \gamma<\xi\}\subseteq Q$ so that
- $s_\gamma\leq t_\gamma\leq s_{\gamma+1}$ for all $\gamma<\xi$ and
- $q_\gamma\Vdash_Q$"if COM plays $\langle \dot p_\beta\mid\beta<\alpha\rangle^\frown\check s_\gamma$ and INC follows $\dot\tau$, INC's final play is $t_\gamma$".
The construction is straightforward and can only break down for two reasons: Either there is no lower bound of $\langle t_\gamma\mid\gamma<\xi\rangle$ or $A_\xi$ is a maximal antichain in $Q$. The latter must happen first since $P$ is ${<}\kappa$-closed and $Q$ is $\kappa$-cc. We set $p_\alpha$ to be the final $s_\xi$ and by mixing of names, we can find a $Q$-name $\dot p_\alpha$ so that
$$q_\gamma\Vdash_Q\dot p_\alpha=\check s_\gamma$$
for all $\gamma<\xi$. This completes the construction.
Finally, let me remark that this is sensitive to the order of play in $G_\kappa(P)$: If $G_\kappa'(P)$ is the game where INC plays at limit steps instead of COM then it is possible that INC has a winning strategy for $G_\kappa'(P)$ in $V^Q$:
Let $Q$ be the poset of finite partial functions $q:\omega_1\rightarrow 2$ ordered by inclusion and let $P=\mathrm{Add}(\omega_1, 1)$. $Q$ is ccc and $P$ is ${<}\omega_1$-closed. Let $G$ be $Q$-generic and consider the strategy for INC for the game $G_\kappa'(P)$ where INC adds a bit of information which agrees with the generic function $\bigcup G$ at the least ordinal which is not yet in the domain of any condition played so far. Assume toward a contradiction that some game in which INC follows this strategy lasts for $\omega_1$-many rounds. The result is a function $f\colon\omega_1\rightarrow 2$ which agrees with $\bigcup G$ on a club $C$. Now $Q$ is ccc, so there is a club $C'\subseteq C$ with $C'\in V$. It follows that $f\notin V$, but we must have $f\upharpoonright\alpha\in V$ for all $\alpha<\omega_1$. This is impossible by a result of Spencer Unger: As $Q\times Q$ is ccc, $Q$ has the $\omega_1$-approximation property, see Lemma 1.2 in Fragility and indestructibility II, Annals of Pure and Applied Logic 166 (2015) 1110-1122.
This is a great question — definitely enjoyed.
Assuming the axiom of choice, then the answer is yes.
Theorem. Assume there is a well ordering of the real numbers. If player I has a winning strategy, then there is a uniform winning strategy.
Proof. Suppose that $\sigma$ is a winning strategy for player I. Fix a well ordering of the collection of sets $B\subseteq\omega$.
Now, suppose that we are playing the game, faced with the actual position $(A_0, x_0, B_0, \ldots,A_n,x_n,B_n)$. But let us look now only at the allowed information for a uniform strategy $(x_0,\ldots,x_n,B_n)$. Consider the least with respect to the well ordering subset $B_n'\subseteq B_n$, such that there is an imaginary play of $B_i'$ that accords with $\sigma$ and the data $(x_0,\ldots,x_n,B_n')$, such that furthermore each $B_i'\subseteq A_i'$ is chosen as least at that stage, subject to being compatible with the data up to $n$. Let $\sigma$ respond to this imaginary play to give us a set $A_{n+1}$, which we now play as the result for this new strategy.
Since the play depended only on $(x_0,\ldots,x_n,B_n)$, I have described a uniform strategy. Let me prove that it is winning. What I claim is that every sequence $x_0,x_1,x_2,\ldots$ arising from a play according to this uniform strategy can be realized as arising in a play according to $\sigma$. The key point is that as $n$ increases, the imaginary sets $B_i'$ will eventually stabilize for each $i$. To see this, consider what happens at stage $n+1$, after we've played $A_{n+1}$. The actual play will respond with some $x_{n+1}\in A_{n+1}$ and $B_{n+1}\subseteq A_{n+1}$, and we will forget about the actual $B_n$ and find some minimal $B_{n+1}''\subset B_{n+1}$ and new imaginary sets $B_i''$ that produce a play according with $\sigma$ and the data $(x_0,\ldots,x_{n+1},B_{n+1}'')$, and is minimal at each stage $i$. But the point now is that the requirement on $B_i''$ is a bit lighter than that on the imaginary set $B_i'$ chosen at stage $n$, provided that $x_{n+1}$ is in $B_n$, which it is. Earlier, we needed $B_i'$ to contain $B_n'$, but now we only need it to contain $B_{n+1}''$, which is strictly smaller. So the set $B_i''$ will be either equal to $B_i'$ or preceed it in the well order. But it can only go down in the well order finitely often, and so eventually, for any given $i$, the imaginary versions of $B_i'''$ will stabilize. In other words, if we have an infinite play according to the uniform strategy, we can find choices of $B_i'''$ that give rise to the whole play. And since this imaginary play accords with $\sigma$, the set $\{x_0,x_1,\ldots\}$ will be in $X$, as desired. $\Box$
Best Answer
What constitutes a partial play doesn't depend on $\alpha$. And $\kappa$ is assumed to have cofinality strictly larger than $\aleph_1$. So he means: define $\sigma$ applied to a partial play to just be some ordinal below $\kappa$ that is above all the outputs of the $\aleph_1$-many strategies $\sigma_\alpha$.
(And it is easy to see that for any fixed $\alpha$, if $\sigma_\alpha$ is a winning strategy for II in the game $G_\alpha$, then so is any function from partial runs into $\kappa$ that is $\ge \sigma_\alpha$.)