Searching for certain compact sets The question asks for a topological space $(X, \tau)$ and a subset $A$ that verifies the following conditions. Can you think of another one for b) and an example for d) with non empty interior?
a) $A$ is not compact and $\bar{A}$ yes: This one is easy, take the real numbers with its usual topology and the interval $(0,1)$.
b) $A$ is compact and $\bar{A}$ isn't: I think that a subset of $\mathbb{R}^n$ is not possible, as the closure of a closed and bounded set is also closed and bounded. Here I have thought of $(\mathbb{N}, \tau)$, where $\tau =\{A_n:n \in \mathbb{N}\}\cup\{\emptyset, \mathbb{N}\}$ and $A_n=\{1, 2, ..., n\}$. Pick $A=\{1\}$. It is compact because it is finite, but the closure is all of $\mathbb{N}$ which is not compact (if it could be covered by a finite number of open sets $\mathbb{N}$ would be open!).
c) $A$ is compact and $\overset{\,\,\circ}A$ isn't: in $\mathbb{R}$ with its usual topology the interval $[0,1]$
d) A isn't compact and $\overset{\,\,\circ}A$ is compact: in $\mathbb{R}$ put $A=\mathbb{Q}$. It isnt compact and its interior (the empty set) trivially is.
Thanks
 A: I'll discuss Part b) first. I think that $\tau$ is not a topology in your example, because it does not include $\mathbb N$ and $\emptyset$, but $\tau\cup\{\emptyset,\mathbb N\}$ is and it is a valid example. As pointed out in the comments, we have to violate the Hausdorff property, so most of the spaces that come to mind won't work. The generalization of your example would be as follows. Let $X$ be a set, and $Y\subseteq X$ such that $Y\neq\emptyset$ and $|\tau|=\infty$, where $\tau=\{U\subseteq X:Y\subseteq U\}\cup\{\emptyset\}$. Then $X$ is the closure of $Y$ because it is the only closed set $C$ with $C\cap Y\neq\emptyset$. Further, we have the cover $X=\bigcup_{x\in X}Y\cup\{x\}$ by open sets which does not admit a finite subcover. It's clearly also not necessary that $Y$ is included in all sets, for example you may also extend this to the disjoint union topology using some other topological space $(X',\tau')$.
For Part d) we start with an easy example, too. Let $B=\mathbb Z_{>0}\times\{1,2\}$, further $B_n=\{(n,2)\}$ and $B'_n=\{(n,3),(n,4)\}$ for $n\in\mathbb Z_{>0}$. Let $\mathcal B=\{B\}\cup\{B_n:n\}\cup \{B'_n:n\}$ and $\tau=\{\bigcup\mathcal A:\mathcal A\subseteq\mathcal B\}$. Notice that $\tau$ is a topology of $X=\mathbb Z_{>0}\times\{1,2,3,4\}$. Let $A=\mathbb Z_{>0}\times\{1,2,3\}$ and notice that $A\subseteq B\cup\bigcup_{n}B'_n$ is a minimal cover, i.e. we cannot omit any of the sets, so $A$ is not compact. Clearly, we have $A^\circ=B$ since this is the only non-empty set contained in $A$. But any open cover of $A$ has to cover $\mathbb Z_{>0}\times\{1\}$, so by the definition of $\tau$ one of the sets in the cover has to be a superset of $A$, so we find a subcover of size $1$. Notice that the $B_n$ are not needed, I included them to illustrate that they don't harm the argument. In general, we would consider $B$, a family $B_i\subseteq B$ of subsets and a family $B'_j$ of sets that are pairwise disjoint and disjoint from $B$, with at least two elements each. So, with $b_i\in B'_i$ we let $A=B\cup\{b_i:i\}$. The remainder is completely analogous to the example.
