This is a great question! I've now managed to eliminate the use of countable choice.
Theorem. Without using any choice principle, it follows that player I can have no winning strategy in the game.
Proof: Working in ZF only, suppose toward contradiction that player I has a winning strategy $\tau$. Look at the set $A$ played by player I on the first move, using strategy $\tau$. Note that $A$ is uncountable, but it can have no uncountable well-orderable subset, since in this case player II could play that subset and move to the well-orderable case, which according to what you have said would put player II in a winning position and therefore contradict that $\tau$ is winning for player I.
Let us now argue that there is a choice function on the uncountable subsets of $A$. To see this, fix any uncountable $B\subset A$ and let us have player II respond with $B$ on player II's first move, with player I continuing subsequently to play according to $\tau$. At each subsequent stage $n$, let us have player II follow the (losing) strategy, where when confronted with a set $E\subset A$, he will find the first rational interval $I$ (in some canonical enumeration) of width $\frac 1n$ such that $E\cap I$ is uncountable, and then play $E\cap I$.
We could deduce that such an interval $I$ exists if we had countable choice, since otherwise $E$ would be a countable union of countable sets. But I claim that we do not need to assume any choice principle in order to make this conclusion. If the set $E$, which was played by player I according to the strategy $\tau$, is the countable union of countable sets $\bigcup_n E_n$, then this is now a winning position for player II, since every uncountable subset of $E$ must contain points from infinitely many $E_n$, and so player II can always subsequently play on his $k^{th}$ step so as to remove the sets $E_i$ for $i\leq k$ from play (a countable removal). Since $\tau$ is winning for player I, however, it must be that the set $E$ is not a countable union of countable sets and so such a rational interval $I$ exists.
Continuing with the main line of argument, what we have is that player II is playing in a prescribed manner (without using any AC), but ensuring that the diameters of the sets in the play of the game go to $0$. Since $\tau$ is winning, the final intersection set will therefore be a singleton subset of $B$, and thus we have provided a means to choose an element from any such $B$.
Using this choice function, we can now choose elements from uncountable subsets of $A$. It follows that we can define an injection of $\omega_1$ into $A$: we first choose a point $a_0\in A$, and then choose a point from $A-\{a_0\}$ and so on. At each stage $\beta$, we choose a point in $A-\{a_\alpha\mid\alpha\lt\beta\}$, which is an uncountable subset of $A$. The set of $a_\alpha$ is an uncountable well-orderable subset of $A$, which contradicts our assumption that there was no such subset. So player I cannot have such a winning strategy after all. QED
This theorem does not seem to imply that player II must have a winning strategy, however, since perhaps the game is undetermined.
Following the link in Andres's comment, we see that Asaf Karagila had had essentially the same argument over at math.SE. And the trick to eliminate the use of countable AC uses the same idea as in Andres's answer over at math.SE.
Here is an amusing concrete non-determined game, under the assumption that the dependent choice principle fails.
Assume DC fails. This means that there is a set $X$ and a binary relation $R$ on $X$, such for every $x\in X$ there is $y$ with $x\mathrel{R} y$, but there is no sequence $\langle x_n\mid n\in\omega\rangle$ with $x_n\mathrel{R} x_{n+1}$ for all $n$. In other words, the tree of finite sequences that accord with $R$ has no leaves — every node can be extended one more step — but there is no way to iterate these steps and the tree has no infinite branch.
Consider the game on $X$ where player I plays a point $a$ from $X$ and player II responds with $b$, and player II wins just in case $a\mathrel{R} b$; otherwise player I wins. So the game is over very quickly, after just one move for each player. (If you insist that games should have infinitely many plays, then you can continue with infinitely many irrelevant moves.)
Player I obviously cannot have a winning strategy, since whatever point $a$ is played, there is a way for II to defeat it by playing some $b$ with $a\mathrel{R} b$.
But player II also cannot have a winning strategy, since any such strategy would provide a choice function on the $R$-successors of the points in $X$, and any such function could be iterated to thread $R$, whereas there is no such infinite thread.
Since the game is an open game (in fact clopen), this argument shows that one cannot prove open determinacy without DC. The usual proof of open determinacy uses DC, when showing that if a position does not have an ordinal rank, then the closed player can avoid losing by maintaining rank. But in order to do so, that player needs DC to find the infinite thread of rank-maintaining moves.
Update. As noted in wojowu's comment below, this example can be adapted to prove the following theorem.
Theorem. ZF proves that there is a non-determined set. Specifically, in ZF we can prove that $\text{AD}_{P(\mathbb{R})}$ fails.
Proof. First, let $F$ be any family of non-empty sets, and consider the game where player I selects some $A\in F$ and player II selects some $a\in A$, and the first player to violate either requirement loses. Clearly, there can be no winning strategy for player I, since if such a strategy directed player I to play a particular $A$ in $F$, then it would be defeated by the strategy for player II to always play some particular element $a\in A$. Conversely, observe that if player II has a winning strategy for the game, then we would have a choice function for $F$. So any violation of AC leads immediately to a non-determined game. In particular, if choice fails for families of nonempty sets of reals, then we get a violation of $\text{AD}_{P(\mathbb{R})}$. On the other hand, if choice holds for such families, then the reals are well-orderable, and we may construct a non-determined set for binary play. So in either case, $\text{AD}_{P(\mathbb{R})}$ fails. QED
Regarding the remarks below about constructing an "explicit" game, one can easily do this as follows.
Corollary. In ZF we can write down an explicit definition of a non-determined set.
Proof. Consider the game where player I chooses either (1) a well-ordering of the reals, and then the rest of the payoff set is as would be constructed for a non-determined set by the usual argument given such a well-ordering; or (2) a family $F$ of nonempty sets of reals having no choice function, with player II then picking an element of the family and player I playing an element of that set, as in the theorem (but with the players reversed). This gives altogether one big payoff set, explicitly defined.
Note that player I cannot have a winning strategy, since he cannot have a strategy for the resulting play in case (1), as the game is undetermined, and he cannot have a strategy that wins by placing him in case (2), since any such strategy would have to provide a choice function for $F$.
Conversely, player II cannot have a winning strategy, since player I can play either a family with no choice function, in which case it is hopeless for II to play with particular element of the family, or else AC holds for all such families in which case player I can play a well-ordering and a non-determined game, for which player II can have no responding strategy. QED
Best Answer
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
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
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.