Since $A$ is compact it is bounded so $A\subseteq B(0,r)$ for some $r>0$.
Then $A\subseteq U\cap B(0,r)$ where $U\cap B(0,r)$ is an open set so by normality of $\mathbb R^n$ an open set $B$ exists with $A\subseteq B\subseteq\overline B\subseteq U\cap B(0,r)\subseteq U$.
Then $\overline B$ is closed and bounded, hence is compact.
Your arguments for (a) and (b) are basically correct, but in places you have badly misused notation. For example, in (b) you write:
So, let $U_\alpha$ be an open cover of $A\cap B$.
So, we have $A\cap B=\bigcup_\alpha U_\alpha$.
If $U_\alpha$ is an open cover of $A\cap B$, as you say in the first sentence, it is a family of open sets. In the second sentence, however, you make it clear that $U_\alpha$ is a single member of some indexed family of open sets. Thus, your first sentence should have read something like this:
So, let $\{U_\alpha:\alpha\in I\}$ be an open cover of $A\cap B$.
Later you write:
$U_\alpha\cup X\setminus A$ is an open cover of $A$
This says that a single set $U_\alpha\cup X\setminus A$ covers $A$. There are two problems here. First, the expression is ambiguous: should it be read $(U_\alpha\cup X)\setminus A$, or should it be read $U_\alpha\cup(X\setminus A)$? Use parentheses when there’s any real possibility of ambiguity. The real problem, though, is that you actually meant something completely different: you meant that $\{U_\alpha:\alpha\in I\}\cup\{X\setminus A\}$ is an open cover of $A$.
By the way, you can avoid indexing altogether. For (a), for instance, let $\mathscr{U}$ be an open cover of $A\cup B$. $A$ and $B$ are compact, and $\mathscr{U}$ is clearly an open cover of each of them, so there are a finite $\mathscr{U}_A\subseteq\mathscr{U}$ that covers $A$ and a finite $\mathscr{U}_B\subseteq\mathscr{U}$ that covers $B$. But then $\mathscr{U}_A\cup\mathscr{U}_B$ is a finite subset of $\mathscr{U}$ that covers $A\cup B$, so $A\cup B$ is compact. Similarly, in (b) you can start with an open cover $\mathscr{U}$ of $A\cap B$. Then $\mathscr{U}\cup\{X\setminus A\}$ is an open cover of the compact set $B$, there is a finite $\mathscr{U}_B\subseteq\mathscr{U}$ such that $\mathscr{U}_B\cup\{X\setminus A\}$ covers $B$. $(A\cap B)\cap(X\setminus A)=\varnothing$, so $\mathscr{U}_B$ covers $A\cap B$, which is therefore compact.
Your answer for (c) is insufficient: you’ve shown that if $X$ is not Hausdorff, the argument that you gave for (b) cannot be carried out, but that does not rule out the possibility that there is some other argument that does not require $X$ to be Hausdorff. To answer (c), you must come up with an actual counterexample. One of the nicest is a convergent sequence with two limits:
Example. Let $X=\Bbb N\cup\{p,q\}$, where $p$ and $q$ are any two points not in $\Bbb N$. Points of $\Bbb N$ are isolated. For each finite $F\subseteq\Bbb N$ let $$B(p,F)=\{p\}\cup(\Bbb N\setminus F)$$ and $$B(q,F)=\{q\}\cup(\Bbb N\setminus F)\,;$$ the sets $B(p,F)$ form a local open nbhd base at $p$, and the sets $B(q,F)$ form a local open nbhd base at $q$. Thus, a set $U\subseteq X$ is open if and only if either $U\subseteq\Bbb N$, or $X\setminus U$ is finite. $X$ is a $T_1$ space and is almost Hausdorff: the only two points that do not have disjoint open nbhds are $p$ and $q$.
Now let $A=X\setminus\{q\}$ and $B=X\setminus\{p\}$; I leave the easy verification that $A$ and $B$ are compact to you. $A\cap B=\Bbb N$, and clear $\big\{\{n\}:n\in\Bbb N\big\}$ is an open cover of $A\cap B$ with no finite subcover.
Best Answer
The 3rd one is wrong because it implies that the union of a collection of closed sets is closed. Suppose every $A\in A'$ is closed. Let $B=\cup_{A\in A'}A.$ Since $\bar A=A$ for every $A\in A',$ we have $\cup_{A\in A'}\bar A=\cup_{A\in A'}A=B$ on the RHS, but the LHS is $\bar B.$
Example. Let $X=\Bbb R$ with the usual topology. Let $B$ be $any$ subset of $\Bbb R.$ Let $A'=\{\,\{r\}: r\in B\}.$
Inferences about finite collections rarely apply to infinite collections.