Why is the choice $\delta_n = 1/n$ so popular in this proof? Yes, another limit question... ;)

I have searched for different proofs of the equivalency of epsilon–delta and sequential continuity. Most sources, e.g., this, that, here, and there, all use the unexplained 'letting' of $$\delta_n=\frac{1}{n}.$$
The theorem is

A function is continuous if and only if it is sequentially continuous.

Let us take this as the example. The final paragraph of the proof is what befuddles me.

Now for each $n \in \Bbb{N^+}$ let $\delta_n=1/n.$ Choose $a_n$ such that $|x-c|<\delta_n=1/n$ and $|f(a_n)-f(c)|\geq\varepsilon_0$. But then we have a contradiction since $(a_n)$ converges to $c$ however $(f(a_n))$ does not converge to $f(c)$. Thus our assumption that $f$ was not continuous at $c$ was false.

'Let $\delta_n=1/n$'? The logic thereafter is understandable, and I am also aware that arriving at the best possible $\delta$ is not necessary, but come on. What is up with this $1/n$? What makes it so ubiquitous?

I do feel like I am supposed to be getting all of this $\varepsilon$–$\delta$ business by now, yet sometimes it cannot be any more frustrating.
 A: A pedantic reason could be due to the Archimedian Property, as I quote from Elementary Analysis by Kenneth A. Ross:

Archimedian Property.
If $a>0$ and $b>0$, then for some positive integer $n$, we have $na>b$
This tells us that, even if $a$ is quite small and $b$ is quite large, some integer multiple of $a$ will exceed $b$. Note that if we set $b=1$, we obtain the following assertion:
If $a>0$, then $\frac{1}{n}<a$ for some positive integer $n$.

Going back to the proof you gave and notice this part:

Choose $a_n$ such that $|a_n - c|<\delta_n = \frac{1}{n}$

The nice thing about using $\delta_n =\frac{1}{n}$ is that you can actually apply Squeeze Theorem because we have both left hand side and right hand side being sequences. As a result of that, it becomes obvious that $\lim_{n\to\infty} |a_n -c|=0$, which really means that $a_n\to c$ as $n\to \infty$.
If, instead, you write

Choose $a_n$ such that $|a_n - c|<\delta$

Then it becomes a little bit tricky, perhaps especially for beginners to realize what really is going on.
A: It is partly because of the answer above (or below), and partly because one can choose $(\delta_n)$ to be any sequence that converges to $0$ for the proof to be valid. Since $(\frac{1}{n})$ is the 'simplest one' of these for most people, and since proving there exists a sequence that converges to $0$ without just explicitly giving one is considered more annoying, people often choose $(\frac{1}{n})$.
