Yes, the closure isn't path connected. To connect a point with $r > 1$ in the set to a point with $r = 1$ in the closure, you'd need a path winding around the unit disk infinitely many times, but that isn't possible since a path is uniformly continuous, and hence a path in the closure can only wind finitely often around the unit disk.
Let us denote $\Gamma = f([\pi/2,+\infty))$. Then $\overline{\Gamma} = \Gamma \cup S^1$ where the union is disjoint. Let $\alpha \colon [0,1] \to \overline{\Gamma}$ be a path with $r_0 := \lVert \alpha(0) \rVert > 1$. Since $\Gamma$ intersects each circle $\{(x,y) : x^2+y^2 = r^2\}$ with $r > 1$ in exactly one point, we can write the polar angle as a function of the radius while $\alpha(t)\in\Gamma$,
$$\varphi(t) = \frac{1}{\log \lVert\alpha(t)\rVert}.$$
Thus to reach the radius $0 < r < r_0$, the polar angle must traverse the entire interval $\left[ \frac{1}{\log r_0}, \frac{1}{\log r}\right]$ and hence $\alpha$ must wind at least
$$n(r) = \left\lfloor \frac{1}{2\pi}\left(\frac{1}{\log r} - \frac{1}{\log r_0}\right)\right\rfloor$$
times around the unit disk. But each full winding around the unit disk contains points where the polar angle differs by $\pi$, and therefore points whose distance is $> 2$.
But, since $[0,1]$ is compact, $\alpha$ is uniformly continuous, so there is a $\delta > 0$ such that for $\lvert t -s\rvert \leqslant \delta$ we have $\lVert \alpha(t) - \alpha(s)\rVert \leqslant 2$. Thus $\alpha$ can wind at most
$$k(\alpha) = \left\lceil \frac{1}{2\delta}\right\rceil$$
times around the unit disk, and therefore
$$n(\lVert\alpha(t)\rVert) < k(\alpha)+1,$$
which implies
$$\lVert \alpha(t)\rVert > \exp \left(\frac{1}{(\log r_0)^{-1} + 2\pi(k(\alpha)+2)}\right) > 1,$$
and hence a path starting in $\Gamma$ never can reach the unit circle.
