I’m afraid that neither argument makes sense. In the first argument it is not true that $x\in\bigcap_{i=1}^kA_i$ iff $x\in A_i$ for all $i\in\Bbb N$. For example, let $A_1=\{0\}$, and let $A_i=\varnothing$ for all $i\ge 2$; then $\{A_i:i\in\Bbb N\}$ is a nest. Now take $k=1$: then
$$\bigcap_{i=1}^kA_i=\bigcap_{i=1}^1A_i=A_1=\{0\}\,,$$
so $0\in\bigcap_{i=1}^kA_i$, but clearly $0\notin A_2$ (and in fact $0\notin A_i$ whenever $i\ge 2$).
In the next line you talk about some $A_j$, but you never define it, so nothing that you say about it is meaningful. At the end you conclude that $x\in A_k$ for all $k\in\Bbb N$; this is just a repetition of what you said in the second line, but with a different name for the subscript, and it is every bit as false in general.
In the argument for (b) you start with an $x\in\bigcup_{i=1}^\infty A_i$ and conclude that it must be in $A_1$, since $A_1\subseteq\bigcup_{i=1}^\infty A_i$. Your reasoning here is exactly backwards: the fact that $A_1\subseteq\bigcup_{i=1}^\infty A_I$ tells you that if $x\in A_1$, then $x\in\bigcup_{i=1}^\infty A_i$, not the reverse.
Then when you try to prove the trivial implication that if $x\in A_i$, then $x\in\bigcup_{i=1}^\infty A_i$, you pick some unspecified positive integers $i$ and $j$ with $i\le j$ and claim that the fact that $A_j\subseteq A_i\subseteq A_1$ somehow proves that $x\in A_i$ for some $i\in\Bbb N$. This is a complete non sequitur, and in any case we already know that $x\in A_i$ for some $i\in\Bbb N$, since we assumed at the beginning that $x\in A_1$, and certainly $1\in\Bbb N$.
To prove (a), first observe that if $x\in\bigcap_{i=1}^kA_i$, then $x\in A_k$, so $\bigcap_{i=1}^kA_i\subseteq A_k$. Now suppose that $x\in A_k$, and let $i\in\{1,\ldots,k\}$; then $i\le k$, so $A_k\subseteq A_i$, and therefore $x\in A_i$. Thus, $x\in A_i$ for all $i\in\{1,\ldots,k\}$, and hence $x\in\bigcap_{i=1}^kA_i$. In other words, $A_k\subseteq\bigcap_{i=1}^kA_i$, and it follows that $\bigcap_{i=1}^kA_i=A_k$.
For (b) it’s immediate that if $x\in A_1$, then $x\in\bigcup_{i=1}^\infty A_i$, so $A_1\subseteq\bigcup_{i=1}^\infty A_i$. Now suppose that $x\in\bigcup_{i=1}^\infty A_i$; then there is some $i\in\Bbb N$ such that $x\in A_i$. And $i\ge 1$, so $A_i\subseteq A_1$, so $x\in A_1$. Thus, $\bigcup_{i=1}^\infty A_i\subseteq A_1$, and we conclude that $\bigcup_{i=1}^\infty A_i=A_1$.
I also disagree with your professor, and think that the new collection of sets will still be pairwise disjoint. But this seems like one of those constructions that are pretty easy to understand conceptually, but are absolute hell to come up with a good notation for.
The down side is that I don't think it's a valid argument, if I understand it correctly as some adaptation of the standard diagonal argument. Because at your induction step you only take a finite number of elements to build your new $A'_{k+1}$! I think you gave yourself an invalid mental picture when you wrote your tableau with row $A'_{k+2}$, and then with "..." for an infinite number of rows. At each inductive step you only have $k$ rows.
You could save the argument by "wrapping the diagonal around" the finite number of rows. I.e. when extracting the values for your new row, after $A_{k,k}$ you go back up to the first row and extract $A_{1,k+1}$, then $A_{2,k+2}$, .... Basically, at step $k$ you take $1/k$ of each existing rows' elements to build your new infinite row. Maybe you had this image in your mind when you used the word 'every' in 'every $ii$-th element', but it didn't come through in your description.
Although once you've fixed it this way, you could see that it could be even easier: At each step $k$, build a new row by taking every other element of your current first row. It's a little easier to describe this procedure, although in both constructions it'd still be a pain to write a formula for what the sequences look like.
Best Answer
Suppose that each $A_i$ is a strict subset of $\mathbb{Z}^+$. Moreover wlog suppose that the sequence of $A_i$ is strictly increasing (pass to such a subsequence otherwise). Then there is a sequence of $a_i$ such that $a_i\in A_i$ and $a_i\notin A_j$ for all $j<i$. This becomes an infinite sequence $B:=\{a_i\}_{i=2}^\infty$. But there is an $n$ such that $B\cap A_n$ is infinite. This is a contradiction since this must contain some $a_m$ with $m>n$.