How was this sequence discovered? Let $N$ be a positive integer and consider the following rational sequence for $n \ge 0$:
$$
a_{n+1} = \frac{N a_n + N}{a_n + N},
a_0 \in \Bbb{Q}.
$$
If $-\sqrt{N} < a_0 < \sqrt{N}$, then $\{a_n\}$ is a monotone increasing rational sequence and converges to $\sqrt{N}$.
If $\sqrt{N} < a_0$, then $\{a_n\}$ is a monotone decreasing rational sequence and converges to $\sqrt{N}$.
By using this sequence, we can easily prove that both $\max\{r |r \lt \sqrt{N}, r \in \Bbb{Q}\}$ and $\min\{r |r \gt \sqrt{N}, r \in \Bbb{Q}\}$ do not exist if N is not a square integer.
This sequence is nice.
How was this sequence discovered?

Bill Trok, thank you very much.
I cannot prove the fact about your sequence but thanks to your answer, I know that the above sequence is not special and I could find another similar sequence.
$$
a_{n+1} = \frac{-4 N^3 -4 N^2 -4 N}{(a_n + 2 N)^2 + 3 N} + N + 1,
a_0 \in \Bbb{Q}.
$$
If $-\sqrt{N} < a_0 < \sqrt{N}$, then $\{a_n\}$ is a monotone increasing rational sequence and converges to $\sqrt{N}$.
If $\sqrt{N} < a_0$, then $\{a_n\}$ is a monotone decreasing rational sequence and converges to $\sqrt{N}$.
http://wolfr.am/1hoilfq
 A: It would not surprise me if it was simply made up for whatever text you found it in. It is not too hard to find other sequences with similar nice properties, and the sequence has the benefit of having a pretty definition. To illustrate this point, you can pick literally any function $f$ where $f(x) > -x$ for all $x > \sqrt{N}$. Then the sequence defined
$$a_{n+1} = \frac{N+a_nf(a_n)}{a_n+f(a_n)}$$
converges to $\sqrt{N}$. Note that your sequence is a special case of this where we have we have $f(a_n) = N$. The proof of this fact is probably very similar to the proof you gave.
One other nice thing I noticed which you might try proving is that for every integer $k > 0$. We can define a sequence $(a_n)$ as follows 
$$a_{n+1} = \frac{N+Na_n}{(a_n)^{k-1}+N}.$$
Then for every $a_0$, where $a_0^k < N$, $(a_n)$ will be a monotonic increasing sequence that converges to $^k\sqrt{N}$. If $a_0$ is positive and $a_0^k > N$, then the sequence $(a_n)$ will be monotonic decreasing and will converge to $^k\sqrt{N}$.
