Describe all the compact subsets of this space Consider the topological space $(X,\mathscr{U})$,  where $X=\mathbb{R}^2$ and the topology $\mathscr{U}$  is generated by the collection of sets $\{(0,0)\}\cup \{I_a\}$ where $I_a$ are the open intervals on the rays departing from the origin.
We then make a quotient of this space, lets call it $Y$, by identifying the points on the closed disk $D^2$ centred at the origin of radius 1.
Intuitively I can see some closed subsets that are compact (essentially closed segments on the rays) and others which are not (any subset extending over a "continuity" of rays). But, being the question "Describe all the compacts", can you suggest me a way to find them all? Is there a procedure someone can follow in this or in similar cases?
Are there some results achievable by finding which separation axiom hold in this space?
Finally, what is the role of the identification in this particular problem? I can't find anything relevant.
 A: The answer provided below should be sufficient to walk you through the proof. I've left some places for you to fill in the proof since this is homework. 
I would start by describing the compact sets in the prequotient space. From here, images of compact sets in the prequotient space are compact in the quotient space (most of the time, these are not the only compact sets). In general, there is not a particularly good method of describing all the compact subsets of a given topological space. As BenjaLim pointed out, the space you described is Hausdorff, so any compact set must be closed. 
Let $(\hat X,\mathscr{U})$ denote the topological space $\hat X=\mathbb{R}^2$ with $\mathscr{U}$ the topology generated by the set $U=\{(0,0)\}\cup\{I_a\}$, where $I_a$ are the open intervals on the rays departing from the origin. Let $(X,\mathscr{V})$ denote the topological space obtained by identifying all the points on the unit disk in the space $(X,\mathscr{U})$. Let $p:\hat X\to X$ denote the quotient map. 
Now what is the space $(\hat X,\mathscr{U})$ like? Well, notice that each open ray from the origin is naturally homeomorphic to $\mathbb{R}$ with the standard topology. So, it is not difficult to see that $(\hat X,\mathscr{U})\cong \coprod_{a\in[0,1]} (X_a,U_a)$, where $\coprod$ denotes the disjoint union space with the disjoint union topology, and the topological spaces $(X_a,U_a)$ are all $\mathbb{R}$ with the standard topology, except $(X_0,U_0)$, which is the one point space. Finally, note that compact sets in the disjoint union topology are precisely those sets which are finite unions $\bigcup_{i=1}^n K_{a_i}$ where $K_{a_i}\subset X_{a_i}$ is compact in the $U_{a_i}$ topology (You should prove this, it's not particularly difficult). So, the compact sets in the prequotient are precisely those which are finite unions of "closed and bounded" (in the subspace topology, which is homeomorphic to $\mathbb{R}$) subsets of the open rays extending from the origin, possibly along with the point $(0,0)$. Hence, the images of these under the map $p$ are all compact by the continuity of $p$. 
Now, what about the quotient space $(X,\mathscr{V})$? The purpose of this identification is to make it so that the space is no longer obviously a disjoint union of simpler spaces. Now, let $b=p((0,1))$, that is, $b$ is the image of the unit disk under $p$. Now, $p$ is a closed map (you could prove this), and so restricting it to the set on which it is injective we get a homeomorphism from $\mathbb{R}^2\setminus\overline{\mathbb{D}}$ (with the subspace topology inherited from $(\hat X,\mathscr{U})$) onto $X\setminus\{b\}$ (with its subspace topology). Thus, any set $K\subset X$ such that $a\notin K$, is compact in $X$ if and only if it is the image under $p$ of a compact subset of $\hat X$. 
From the previous paragraph, we have reduced the problem to describing the compact sets of $X$ that contain the point $a$. Let $K\subset X$ be compact with $a\in K$. Suppose that there exist points $a_n\in K$ with $a_n\not=a$ such that $\hat{a}_n=p^{-1}(a_n)$ have the property that $\hat{a}_n$ and $\hat{a}_m$ lie on different rays if $m\not=n$. Now we form saturated open sets $\hat{V}_n$ in $\hat X$ such that $\hat{a}_k\in \hat{V}_n$ if and only if $k\le n$, and that $\bigcup \hat{V}_n = \hat X$. Then $K\subset\bigcup V_n$, but is not contained in any finite subcollection, which will contradict the compactness of $K$. Hence $K$ is the image of a set that is contained in finitely many of the rays extending from the origin. I leave it to you to construct some appropriate $\hat{V}_n$ sets (I know these exist, I have constructed them). 
The previous paragraph proves that any compact set in $X$ must be the image of a compact set in $\hat{X}$. This completes the proof. 
