To be able to generalize procedures across various epsilon-delta proofs, it is important to notice what are the stand out features of such proofs(tricks, conversions etc.)
In this case, suppose we want to show that $L=0$. Let us take $\epsilon>0$. We want to find an $N >0 $ such that if $n > N$ then $|a_n - L| < \epsilon$. In our case, $L=0$, so it changes to $|a_n| < \epsilon$.
Step 1: Look at $a_n$ carefully. It is $\dfrac{(-1)^{n+1}}{n}$. Now, we are required to find out about $|a_n|$, so let us compute $|a_n|$. This is a step by itself, because it gives a clear direction of attack: To attack this problem, we will calculate $|a_n|$ explicitly, and try to find $N$ satisfying the limit conditions explicitly. We normally do this because $a_n$ is not a very complicated quantity, so it is easy to work with.
Step 2:So what is $|a_n|$? It is $\left|\dfrac{(-1)^{n+1}}{n} \right|$. Since the modulus splits across the fraction,
$$
\left|\dfrac{(-1)^{n+1}}{n} \right| = \dfrac{|(-1)^{n+1}|}{|n|} = \dfrac{1}{n}
$$
There is no trick here. We just simply calculated $|a_n-L|$ directly, because in this problem it was easy to do so. The reason for this step is motivated by the previous step.
Step 3: Now, suppose we were given $\epsilon>0$ and were asked to find a large enough $N$ such that if $n > N$ then $|a_n| < \epsilon$. However, we have now calculated $|a_n|$, and it is $\frac 1n$. Hence, we are trying to find $n$ such that $\frac{1}{n} < \epsilon$. However:
$$
\epsilon > \frac 1n \iff n > \frac{1}{\epsilon}
$$
The above piece of insight is vital to us: we can find our $N$ explicitly.
Step 4: Let $N$ be the smallest integer greater than $\frac{1}{\epsilon}$. Then, note that:
$$
n > N \implies n > \frac 1 \epsilon \implies \epsilon > \frac 1n
$$
Step 5: Hence, this $N$ works for the given problem, so we can conclude by definition of limit that $a_n \to 0$.
What have we learnt from here?
1) Wherever possible and easy, calculate $|a_n -L |$ explicitly. It is the safest option for simple looking $a_n$.
2) In Step $3$, we actually worked backwards. We assumed that the $N$ which we wanted existed, and then we tracked back to actually find that $N$, right? Working backwards is a very big trick, because the $N$ that you want can be explicitly found via working backward, as has happened in this case.
Unfortunately, the problem that you have is not very illustrative, because it doesn't go through all the tricks and twists that one goes through while evaluating a normal tricky limit. You need to find a good example (and me), and then you will get a better grip on limits, because (certainly unknown to you, so please don't berate yourself) this problem was damp squib compared to some of the harder limits you have to come across. Hopefully, though, I have done justice to this problem. Please get back on doubts.
First we need to note that the notation in the question makes no sense: If $f:\Bbb N^2\to\Bbb N$ is a bijection and $(x_{n,m})$ is a double sequence then " $\{x_{n,m}\}$ converges if and only if $\{x_{f(n,m)}\}$ converges" is impossible. Because there's simply no such thing as $x_{f(m,n)}$, since $f(m,n)\notin\Bbb N^2$.
There are various ways to state the natural question that seems likely to be what the OP really meant to ask about, for example
Q. Does there exist a bijection $f:\Bbb N \to\Bbb N^2$ such that for any doubly-indexed sequence $x_{m,n}$ we have $x_{m,n}\to x$ (as defined above) if and only if $x_{f(k)}\to x$?
The answer is no, there is no such bijection. Given $f$, choose $k_n\in\Bbb N$ with $$f(k_n)=(0,n).$$
Easy Exercise. $\lim_{n\to\infty}k_n=\infty$.
Now let $$x_{m,n}=\begin{cases} 1,&(m=0),\\0,&(m>0).\end{cases}$$ Then $x_{m,n}\to0$. But $x_{f(k_n)}=1\not\to0,$ and since $k_n\to\infty$ this shows that $$x_{f(k)}\not\to0.$$
Best Answer
Let us prove the convergence of $a_n$ to 1.
Claim : For all $\epsilon\gt 0\ \exists\ N(\epsilon) \in\mathbb{N}$ such that for all $n\in\mathbb{N}, n\ge N$,
$|a_n-1|<\epsilon$
Proof : Fix an $\epsilon \gt 0$
We note that $a_n$ is increasing, and $\forall n, a_n\lt 1$.
Therefore if for some $N(\epsilon),1-a_N\lt\epsilon$, then for all $n\ge N,1-a_n\lt\epsilon\;$ (We remove the modulus because $a_n\lt 1$)
Let us find such an $N$. $$1-a_N\lt\epsilon\Rightarrow \frac1{N^2+1}\lt\epsilon\Rightarrow N^2\gt\frac 1\epsilon-1$$
If $\frac 1\epsilon -1\lt 0$, then the inequality is true for all $N$. If $\frac 1\epsilon -1\ge 0$, then $N$ should be greater than $\sqrt{\frac 1\epsilon -1}$.
That's it. For an arbitrary $\epsilon\gt 0$, we found an $N(\epsilon) \in\mathbb{N}$ such that for all $n\in\mathbb{N}, n\ge N$,
$|a_n-1|<\epsilon$, and so $\lim_\limits{n\to\infty}a_n=1$