Thanks for pointing out the potential different meanings of CG and the dependence on context. In the example at hand, the starting space is a sequential space (closed sets are precisely those closed under convergent sequences) and convergent sequences have unique limits.

$Y$ is compact and hence CG. $X$ is $T_2$ and hence compact subsets of $X$ are closed in $X$. Thus $Y$ is the Alexandroff compactification of $X$. In particular $Y$ is compact and infinite and contains $y$.

If $X$ is metric then $X$ contains a disk iff there exists a map from $R^{2}$ onto a subspace $Y$ of $X$, such that compacta in $Y$ have compact preimage, and so that the preimage of each $y$ in $Y$ does not separate the plane.

Iff there exists a quotient map from $R^{2}$ onto a subspace of $X$ such that each point preimage is a non separating continuum, and so that the point preimages form an upper-semi-continuous decomposition of the plane. There is some redundancy in this answer if $X$ has a reasonable topology.

Nice question! If forced to bet, I would bet no, despite having no counterexample in mind,

As noted, thanks to Steven's inquiry in the comments of the original question, I noticed the mistake in my original answer and corrected it yesterday, and, serendipitously, also constructed a non-example similar to the one at hand. It's nice to revisit the question, thanks to Steven and everyone else for exposing the mistake, and helping to set things right.

To see why we must also assume $X$ is compactly generated, let $X$ be the plane, but refine the usual topology so that each countable set is closed. Then no infinite subset of $X$ is compact. However if $Y$ is the Alexandroff compactifictation then each subset of $Y$ which contains $\infty$ is compact. In particular $Y \setminus 0$ is compact but not closed in $Y$.

Thanks Steven, nice catch! We should also assume $X$ is compactly generated, i.e. $A$ is closed in $X$ iff $A \cap C$ is closed in $X$ for each compact $C \subset X$.

But isn't $Z_{p}$ uncountable and consequently NOT cyclic? ncatlab.org/nlab/show/p-adic+integer. Treated as a group, aren't the units of its associated ring extension precisely the group automorphisms of $Z_{p}$? Please see Andreas Blass's answer below.

The comment two doors back is not correct. If the component A of the decomposition of the closed unit disk contains a two cell, then, A also has 2 or 3 semicircles from the upper half disk attached to A.

We can demand a bit more geometry from our decomposition. Ignoring the upper unit disk, each point x on the lower semicircle belongs to a line segment connecting the horizontal bar, or a hyperbolic triangle with two straight sides, and two points on the horizontal bar.

For the remaining points x in the open upper unit disk, x belongs to a semcircle union an arc in the lower closed unit disk, or a semicircle union a 2-cell in the lower closed unit disk. There are a handful of other exceptions. So far all the mentioned pieces are cellular.

Given the constructed decomposition of the closed unit disk, a random point in the open upper unit disk belongs to a semicircle.