For those who don't like spirals, another trick is to express the points as complex numbers $$x \,=\, r e^{i \varphi}$$. All the points on the line perpendicular to the radial passing through $$x$$ can then be expressed in the form $$( 1 + i \varepsilon ) x$$ for some parameter $$\varepsilon$$. So if Line always uses this strategy we can cast the problem as analysing the convergence of expressions like:
$$
x_{n} \,=\, \prod_{k=1}^{n} ( 1 \,+\, i \varepsilon_{k} ) \, x_{1}
$$
In norm this is:
$$
| x_{n} |^{2} \,=\, \prod_{k=1}^{n} ( 1 \,+\, \varepsilon_{k}^{\text{ }2} ) \, | x_{1} |^{2}
$$
which puts the only constraint on the $$\varepsilon_{k}$$'s when we require that the sequence converges to a point within the circle.
One possibility is to set $$\varepsilon_{k} \,=\, \frac{\tilde{\varepsilon}}{n}$$, so that:
$$
x_{n} \,=\, \bigl( 1 \,+\, \frac{i \tilde{\varepsilon}}{n} \bigr)^{n} x_{1}
$$
For sequences such as this, this becomes a phase change as we let $$n \rightarrow \infty$$:
$$
\lim_{n \rightarrow \infty} \bigl( 1 \,+\, \frac{i \tilde{\varepsilon}}{n} \bigr)^{n} \,=\, e^{i \tilde{\varepsilon}}
$$
To show what's already been demonstrated in this thread, while Point can't do "infinitely many" "infinitesimal" steps, he can still decide his next $$n$$ steps will be $$x_{k+1} \,=\, \bigl( 1 \,+\, \frac{i \tilde{\varepsilon}}{n} \bigr) x_{k}$$ for arbitrarily large $$n$$, and so he can increase his phase by any desired amount with an arbitrarily small increase in radius (ie. he can get arbitrarily close to the limiting case of a pure phase change of arbitrary phase $$\tilde{\varepsilon}$$). Since Point has this possibility at any stage of the the game, Line can't force Point to converge in this way.
