Proving $\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k \subseteq \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$, and a condition for equality Define $B:=\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k$ and $C:=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$
Show:
$B \subset C$
And if
$\forall k \in \mathbb{N}_+:A_k \subset A_{k+1}$ or $\forall k \in \mathbb{N}_+: A_{k+1}\subset A_k$ then $B=C$.
This is what I did for the first question:
Pick $x \in B$, then $\exists n \in \mathbb{N}_+:\forall k\ge n:x \in A_k$. So set $n=n^*$ and pick $k_0\ge n^*$ so, that $x \in A_k$.
Suppose $x \not \in C$, then $\exists n \in \mathbb{N}_+ \forall k \ge n: x \not \in A_k$, but this contradicts the fact above; so $x \in C$. Is this OK and how to do the second question?
 A: First, there is no need to argue by contradiction:
Let $x\in B$. Then there exists $N$ such that $x\in \cap_{k=N}^{\infty} A_k$. In particular, for all $j\geq N$, $x\in A_j$. Therefore, since for every $n$ there exists $k$ such that $k\geq\max{N,n}$, we have $x\in \cup_{k=n}^{\infty}A_k$ for all $n$, so $x\in \cap_{n=1}^{\infty}(\cup_{k=n}^{\infty}A_k = C$. Thus, $B\subseteq C$.
Also: you never actually use $k_0$ (I assume it's a typo and you want $x\in A_{k_0}$ in the penultimate paragraph). But the main idea is correct, and the argument (once the typo is corrected) is fine.
For the second question: since $B\subseteq C$ holds with no conditions, we only need to show that under either of the given conditions we would have $C\subseteq B$. We can argue by contrapositive: assume there exists $y\in C$ such that $y\notin B$, and show that the negation of the condition holds; the negation is:
$$\left(\exists k\in\mathbb{N}_+\text{ such that } A_k\not\subset A_{k+1}\right)\text{ and }\left(\exists \ell\in\mathbb{N}_+\text{ such that } A_{\ell+1}\not\subset A_{\ell}\right).$$
Now, $y\in C$, so for every $n$ there exists $m\geq n$ such that $y\in A_m$. But $y\notin B$, so for every $n$ there exists $m'\geq n$ such that $y\notin A_m$. 
Pick $n=1$; then there is an $N\geq 1$ such that $y\in A_N$; and there is an $M\geq N$ such that $y\notin A_M$ (note that in fact we must have $M\gt N$). Now, since $y\in A_N$ but $y\notin A_M$, there is a value of $k$, $N\leq k\lt M$, such that $y\in A_k$ but $y\notin A_{k+1}$. For that value of $k$, we have $A_k\not\subseteq A_{k+1}$. That establishes the first clause of the conjunction we need to establish.
I'll let you establish the second clause. 
