How does exactly the $\epsilon$-$\delta$ method tells me I am right? Sometimes, tiny little needles hit me under the form of perhaps stupid questions which I need to ask to clear them. This times it's about $\epsilon$-$\delta$ proof for limits.
I know the definition, but I need a clarification on this example.
The definition I know is:

$$\lim_{x\to p} f(x) = L$$ iff $\forall \epsilon > 0$, there exists $\delta > 0$ such that for all $0 < |x - p| < \delta$ we have $|f(x) - L| < \epsilon$.

Ok, the definition might be in contrast to my procedure but here is also the "strategy" I understood to run:
I first solve $|f(x) - L| < \epsilon$ for $x$. Once I know the values for $x$ that will work, I choose $\delta$ such that the interval $(a - \delta, a + \delta$ sits inside the steps of solutions. But to find $\delta$ I must first solve and find the range for $epsilon$ (when I solve for $x$). Here an example:
Be
$$\lim_{x\to 1} x^2+1 = 2$$
which is of course true, but I will verify it through $\epsilon-\delta$: the passages are the following
$\bullet -\epsilon < x^2+1-2 < \epsilon$
$\bullet 1-\epsilon < x^2 < 1 + \epsilon $
$\bullet \sqrt{1-\epsilon} < x < \sqrt{1+\epsilon}$
Now, since $2$ is the limit, those are the other passages for the $\delta$:
$\bullet 2 - \delta > \sqrt{1-\epsilon}$
$\bullet 2 + \delta < \sqrt{1 + \epsilon}$
In the end
$$\delta = \text{min}\left(2 - \sqrt{1 - \epsilon}, \sqrt{1+\epsilon}-2\right)$$
Fine.
The point now is: suppose I say the limit (the same above) makes $5$ and not $2$.
All the "same" passages will lead to
$$\delta = \text{min}\left(6 - \sqrt{5 - \epsilon}, \sqrt{5 + \epsilon} - 6\right)$$
The question
How and why the last expression for $\delta$ tells me I am wrong, and the limit is not $5$?
LET ME SEE IF I PERHAPS UNDERSTOOD
After a bit of reading and thinking, I think I came to a solution for my problem.
Redoing the whole procedure, I have:
$\bullet -\epsilon < x^2 + 1 - 2 < \epsilon $
$\bullet \sqrt{1-\epsilon} < x < \sqrt{1 + \epsilon}$
At this point I need to find a $\delta$ such that $|x - 1| < \delta$
I then subtract $1$:
$$\sqrt{1 - \epsilon} - 1 < x - 1 < \sqrt{1 + \epsilon} - 1$$
A problem yet persists: how could I treat this? I may find a better estimation, like $\sqrt{1 + \epsilon} < 1 + \epsilon$ hence
$-\epsilon < x-1 < \epsilon$ which makes me to take $\delta = \epsilon$
Is this right?
 A: Here is the standard way to do it.
$|(x^2 + 1) - 2 | < \epsilon\\
|x^2 - 1| < \epsilon\\
|(x+1)(x-1)| < \epsilon$
Let's say that $\delta \le 1$
$|x-1| < \delta \implies |x+1| < 3$
$|(x+1)(x-1)| < 3\delta \le \epsilon$
let $\delta = \min (1,\frac \epsilon 3)$
As for your work above.
$\bullet 2 - \delta > \sqrt{1-\epsilon}\\
\bullet 2 + \delta > \sqrt{1+\epsilon}$
Seems to be coming out of nowhere.
A: You have put it correctly in an update to the question:

At this point I need to find a $\delta$ such that $|x - 1| < \delta$
I then subtract $1$:
$$\sqrt{1 - \epsilon} - 1 < x - 1 < \sqrt{1 + \epsilon} - 1.$$

But here's the reason why you wouldn't want to replace $\sqrt{1 + \epsilon}$ by
$1 + \epsilon$ in this inequality: it's precisely because
$\sqrt{1 + \epsilon} < 1 + \epsilon,$ which means that by making this replacement you allow a greater upper bound of $x$ and you therefore allow $x$ to take on values that $\sqrt{1 - \epsilon} - 1 < x - 1 < \sqrt{1 + \epsilon} - 1$
would not allow it to take.
But $\sqrt{1 - \epsilon} - 1 < x - 1 < \sqrt{1 + \epsilon} - 1$
was carefully crafted to give you exactly the set of $x$ for which
$\lvert f(x) - L\rvert < \epsilon$ in your example;
as soon as you let any other value of $x$ be possible, you are allowing $x$ to take at least one value such that $\lvert f(x) - L\rvert \geq \epsilon$
and you will no longer have a proof.

As an aside, I don't particularly like the strategy that says you must deduce the largest possible value of $\delta$ that you can find that will still allow the proof to work (as strategies like the above seem designed to do).
But you very well can replace expressions like $\sqrt{1+\epsilon}$
with something simpler if you like;
the thing is just that you can only lower the upper bound and only raise the lower bound.
To put it another way,
"$\delta$ can never be too small."
So I might work it in this direction instead:
I am going to pick some value of $\delta$ (I haven't decided which, yet),
after which I'm going to prove something implied by
$$ 0 < \lvert x - 1\rvert < \delta. $$
I know that if I ensure $\lvert x - 1\rvert < 1$ (which I can do just by making sure that the value of $\delta$ I choose is not greater than $1$),
then $x$ is in the open interval $(0, 2)$,
$x^2 + 1$ is increasing over the whole interval, and the worst-case scenario for $\lvert f(x) - 2\rvert$ is either at $x = 1 + \delta$ or $x = 1 - \delta.$
Specifically,
$$ \lvert f(1 + \delta) - 2\rvert = (1 + \delta)^2 + 1 - 2 = 2\delta + \delta^2 $$
and
$$ \lvert f(1 - \delta) - 2\rvert = 2 - ((1 - \delta)^2 + 1) = 2\delta - \delta^2. $$
Comparing the two, it's clear the worst case (the greatest $\lvert f(x) - 2\rvert$)
is $2\delta + \delta^2,$
so I just need to ensure that
$$ 2\delta + \delta^2 < \epsilon. $$
Since I already decided I would not choose $\delta > 1,$
I can be sure that $2 + \delta \leq 3$.
So rather than trying to solve the a quadratic equation in $\delta$ I'm just going to try to set it up so that
$$ 2\delta + \delta^2 = (2 + \delta)\delta \leq 3\delta < \epsilon. $$
Let's try $\delta = \min\left(\frac14 \epsilon, 1\right)$;
the $1$ comes from the fact that I already said I will not let $\delta$ be greater than $1,$ and the $\frac14 \epsilon$ is a relatively simple way of ensuring that
$3\delta < \epsilon.$
Now it might seem I made $\delta$ smaller than necessary. 
And I did make $\delta$ smaller than necessary; I don't care.
The definition of the limit merely says there has to exist a $\delta$ satisfying certain conditions; it makes no requirement for me to tell you what the largest possible $\delta$ value is. 
It only matters that you will be able to prove that if
$1 - \delta < x < 1$ or $1 < x < 1 + \delta$
then $-\epsilon < (x^2 + 1) - 2 < \epsilon.$
