Hint: (i)For $A \cup B$, when you have an open cover for $A \cup B$, you have an open cover for both $A$ and $B$ and by compactness of $A$ and $B$, you have a finite subcover for $A$ and a finite subcover for $B$ and hence a finite subcover of $A \cup B$.
(ii) For $A \cap B$, $A$ and$B$ are compact in a Hausdroff space. Are they closed? Is $A \cap B$ closed? What can you say about a closed subspace of a compact space? Is it compact again?
Another answer to address the proof the OP proposed.
To set the stage:
So we have the set $X=\Omega \cup \{\Omega\}$ where $\Omega$ is the smallest uncountable ordinal. $\Omega \in X$ is the maximum of $X$. $X$ has the order topology, which has as a base all sets of the following forms
- $[0,x), x \in X$ (the only basic open sets that contain $0$),
- $(x,y): x < y \in X$,
- $(x,\Omega], x \in X$ (the only basic open sets that contain $\Omega$).
Now let $\mathcal{U}$ be an open cover of $X$. Suppose that it does not have a finite subcover (striving for a contradiction). Define $$M = \{x: [0,x] \text{ does not have a finite subcover by } \mathcal{U}\}$$
and note that, as $\Omega \in M$, $ M \neq \emptyset$, so $m=\min(M) \in X$ exists. Also $0 \notin M$, clearly.
First case: $m=\Omega$. Then as we have a cover of $X$ some $U_0 \in \mathcal{U}$ exists with $\Omega \in U_0$, and so there is some $x_0 \in X$ such that $(x_0,\Omega] \subseteq U_0$. As $x_0 < m=\min(M)$, $[0,x_0]$ has a finite subcover by elements from $\mathcal{U}$ (else it would contradict the minimality of $m$) and adding $U_0$ to these, shows that we also have a finite subcover for $[0,m]$, a contradiction, as then $m \notin M$ at all.
Second case (really the same) $m < \Omega$ (or equivalently $0< m \in \Omega$): then again we find $U_0$ from $\mathcal{U}$ such that $m \in U_0$ and a basic $(x_0, x_1)$ with $m \in (x_0, x_1) \subseteq U_0$. Again we have a finite subcover for $[0,x_0]$ and we add $U_0$ to cover $[0,m]$ again by a finite subcover and so $m \notin M$ contradiction.
We could have combined the above two cases by a priori showing that in a well-order the order topology has as a base all sets of the form $(x,y], x < y \in X$ and use only sets of that form in our argument.
But the minimality argument is a valid one. The max is needed for non-emptyness of $M$, that's why the argument fails for the non-compact space $\Omega$ itself. This clearly has a cover $\{[0,x): x \in \Omega\}$ without even a countable subcover (as all countable subsets of $\Omega$ have an upperbound in $\Omega$).
Best Answer
Since the collection $\mathcal{C}$ is an open cover for $A$ and $\ell \in A$, there must be some open set $U_0 \in \mathcal{C}$ such that $\ell \in U_0$. Since $U_0$ is open and $\ell \in U_0$, there is $\varepsilon>0$ so that $$(\ell-\varepsilon,\ell+\varepsilon) \subseteq U_0.$$ Since $\lim_{n\to \infty} a_n=\ell$, there is $N \in \mathbb{N}$ so that $$|a_n-\ell|< \varepsilon \text{ whenever } n \geq N.$$ Now there are at most finitely many ($N-1$ or $N$ depending if you include $0\in \mathbb{N}$) elements in $A$ that are not in $U_0$ but since $\mathcal{C}$ is a cover this just means we choose a set from $\mathcal{C}$ for each one of these finitely many terms at the start of the sequence $\{a_n\}$.
Hope this helped but not too much.