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Here is a sketch of a proof building on the ideas of Andreas Blass.

Let $C(r)$ be the circle of radius $r$ centered at the origin $O$. Suppose that $P$ starts at $p \in C(r)$ for some $r \in (0,1]$. To begin, let us analyze what happens if $P$ follows the simple strategy of always staying on $C(r)$. That is, after $L$ chooses a line $l$ through $p$, $P$ picks the other point of $l$ that is also on $C(r)$ (if $l$ happens to be the tangent to $C(r)$ at $p$, then $P$ picks $p$). This is clearly not a winning strategy for $P$. For example, $L$ could force $P$ to always stay at $p$ by always picking the tangent to $C(r)$ at $p$. However, if $(p_i)_{i=1}^\infty$ is the set of chosen points and if $\theta_i$ is the smaller of the two angles made by $p_{i}, O$, and $p_{i+1}$, then $P$ will win if the $\theta_i$ do not tend to 0.

Returning to the original game, the second observation is that if $L$ follows the strategy of always choosing tangent lines, then $P$ has a winning strategy. This strategy was suggested by Gerhard Paseman in the comments. The fact that $P$ has a winning strategy follows by Andreas' answer, because in this case there are always two choices for $p_n$, and so we can always move clockwise when jumping from $C(r_n)$ to $C(r_{n+1})$.

Putting these two observations together yields a winning strategy for $P$ as follows. As in Andreas' answer, fix a series $\sum_1^\infty t_n$ of positive terms, with sum 1, such that $\sum_1^\infty \sqrt{t_n}$ diverges. Let $r_n=\sum_1^n t_k$. Inductively assume that $P$ is currently on $p \in C(r_n)$. At this point, $P$ follows the strategy of staying on $C(r_n)$. Recall that this is winning for $P$ unless $\theta_i \to 0$. So, $P$ simply needs to wait until $\theta_i$ is sufficiently small at which point he jumps to $C(r_{n+1})$ in the clockwise direction.

Edit. In light of Ravi Boppana's answer, I see an error (there must be) in the above argument. While $P$ is waiting on $C(r_n)$, it is true that $L$ must eventually pick a line that is arbitrarily close to being a tangent. However, $P$ has no control when this will happen. In particular, there is no guarantee that the sequence $(p_i)_{i=1}^{\infty}$, where $p_i$ is the first point chosen by $P$ on $C(r_i)$, diverges.

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Here is a sketch of a proof building on the ideas of Andreas Blass.

Let $C(r)$ be the circle of radius $r$ centered at the origin $O$. Suppose that $P$ starts at $p \in C(r)$ for some $r \in (0,1]$. To begin, let us analyze what happens if $P$ follows the simple strategy of always staying on $C(r)$. That is, after $L$ chooses a line $l$ through $p$, $P$ picks the other point of $l$ that is also on $C(r)$ (if $l$ happens to be the tangent to $C(r)$ at $x$, p$, then$P$picks$x$). p$). This is clearly not a winning strategy for $P$. Letting For example, $L$ could force $P$ to always stay at $p$ by always picking the tangent to $C(r)$ at $p$. However, if $(p_i)_{i=1}^\infty$ be is the set of chosen points and letting if $\theta_i$ be is the smaller of the two angles made by $p_{i}, O$, and $p_{i+1}$, we see that then $L$ wins if and only P$will win if the$\theta_i \\theta_i$do not tend to 0$0.

The

Returning to the original game, the second observation is that if $L$ follows the strategy of always choosing tangent lines, then $P$ has a winning strategy. This strategy was suggested by Gerhard Paseman in the comments, then . The fact that $P$ has a winning strategy . This follows by Andreas' answer, because in this case there are always two choices for $p_n$, and so we can always move clockwise when jumping from $C(r_n)$ to $C(r_{n+1})$.

Putting these two observations together yields a winning strategy for $P$ as follows. As in Andreas' answer, fix a series $\sum_1^\infty t_n$ of positive terms, with sum 1, such that $\sum_1^\infty \sqrt{t_n}$ diverges. Let $r_n=\sum_1^n t_k$. Inductively assume that $P$ is currently on $p \in C(r_n)$. At this point, $P$ follows the strategy of staying on $C(r_n)$. Recall that this is winning for $P$ unless $\theta_i \to 0$. So, $P$ simply needs to wait until $\theta_i$ is sufficiently small at which point he jumps to $C(r_{n+1})$ in the clockwise direction.

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Here is a sketch of a proof building on the ideas of Andreas Blass.

Let $C(r)$ be the circle of radius $r$ centered at the origin $O$. Suppose that $P$ starts at $p \in C(r)$ for some $r \in (0,1]$. To begin, let us analyze what happens if $P$ follows the simple strategy of always staying on $C(r)$. That is, after $L$ chooses a line $l$ through $p$, $P$ picks the other point of $l$ that is also on $C(r)$ (if $l$ happens to be the tangent to $C(r)$ at $x$, then $P$ picks $x$). This is clearly not a winning strategy for $P$. Letting $(p_i)_{i=1}^\infty$ be the set of chosen points and letting $\theta_i$ be the smaller of the two angles made by $p_{i}, O$, and $p_{i+1}$, we see that $L$ wins if and only if $\theta_i \to 0$.

The second observation is that if $L$ follows the strategy suggested by Gerhard Paseman in the comments, then $P$ has a winning strategy. This follows by Andreas' answer, because in this case there are always two choices for $p_n$, and so we can always move clockwise when jumping from $C(r_n)$ to $C(r_{n+1})$.

Putting these two observations together yields a winning strategy for $P$ as follows. As in Andreas' answer, fix a series $\sum_1^\infty t_n$ of positive terms, with sum 1, such that $\sum_1^\infty \sqrt{t_n}$ diverges. Let $r_n=\sum_1^n t_k$. Inductively assume that $P$ is currently on $p \in C(r_n)$. At this point, $P$ follows the strategy of staying on $C(r_n)$. Recall that this is winning for $P$ unless $\theta_i \to 0$. So, $P$ simply needs to wait until $\theta_i$ is sufficiently small at which point he jumps to $C(r_{n+1})$ in the clockwise direction.