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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 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 $u_{\ast }$ be its first contact with the disk. The disk is closed, so this first contact exists, and $u_{\ast }>0$. Before it, the path belongs to $Y$ and has a continuous uniquely determined parameterContinuity at contact forces $|\alpha (u_{\ast })|=1$, and hence $t(u)\to \infty$ as $u\uparrow u_{\ast }$. The intermediate value theorem supplies times approaching $u_{\ast }$ 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 $\alpha (u)\to 1$; along the second it tends to $-1$. This contradicts continuity at $u_{\ast }$. Thus no path joins the spiral to the disk, proving failure of path-connectedness for the spiral accumulating on a circle.