= Solution
The set $Y$ is <path-connected> as the continuous image of $[1,\infty)$, hence connected. Its closure is $Y$ together with the unit circle. Indeed $g(k+\theta/(2\pi))\to e^{i\theta}$ as $k\to\infty$, and any convergent sequence of spiral points either has bounded parameters, giving a point of $Y$, or has parameters tending to infinity along a subsequence, giving radius one. The closure of a connected set is connected: a separation of its closure would induce a separation of the dense original set, with neither open part missing that set. The closed disk is connected and meets this closure in the unit circle; their union $Y\cup\overline D$ is therefore \b[connected].
To show it is not <path-connected>, suppose a path $\alpha:[0,1]\to Y\cup\overline D$ starts in $Y$ and ends in the disk. Let $u_*$ be its first contact with the disk. The disk is closed, so this first contact exists, and $u_*>0$. Before it, the path belongs to $Y$ and has a continuous uniquely determined parameter
$$
t(u)=\frac1{|\alpha(u)|-1},\qquad\alpha(u)=g(t(u)).
$$
<Continuity> at contact forces $|\alpha(u_*)|=1$, and hence $t(u)\to\infty$ as $u\uparrow u_*$. The <intermediate value theorem> supplies times approaching $u_*$ at which $t(u)$ is an integer, and other times at which it is an integer plus $1/2$. These times must approach contact because $t$ is bounded on every compact interval strictly before it. Along the first sequence $\alpha(u)\to1$; along the second it tends to $-1$. This contradicts <continuity> at $u_*$. Thus \b[no path joins the spiral to the disk], proving failure of path-<connectedness> for the <spiral accumulating on a circle>.
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