How to find the desired Epsilon? Say for example I want to proof that:
$\forall 0<c \in \mathbb{Q} \; $ st. $\;c^2 < 2 \quad \exists 0<\varepsilon \in \mathbb{Q}\;$ st. $\; (c + \varepsilon)^2 < 2 $
And moreover, I want to actually find one.
Iv'e already been convinced it works with $\ \varepsilon = \frac{2-c^2}{c+2} \;$,
Now as I see it, it is just a way of dividing the distance between $c^2$ and $2$, but I'm wondering about  the characteristic of this quotient.
My first question is how could i have found this, or any other $\ \varepsilon \; $ that satisfies ?
I was told that by solving $\; (c + \varepsilon)^2 < 2 $ for $\ \varepsilon \; $ I could have found one, but I can't really see how it works with the demand of it being a rational number (?). Now this is only one example, and finding a desired element is something I have seen repeats many times. Are there any main guidelines for this?
BTW,
I'm very new here and this is my very first post.

Let $0<\varepsilon<1$ be a small number; then we have $$(c+\varepsilon)^2=c^2+2\varepsilon c+\varepsilon^2\le c^2+4\varepsilon+\varepsilon=c^2+5\varepsilon$$ since $c\le2$ and $\varepsilon^2\le\varepsilon$. Since $c^2<2$, we see that we can choose an $0<\varepsilon<1$ such that $c^2+5\varepsilon<2$, thus $(c+\varepsilon)^2<2$.
We can prove (by Archimedean property) that

Lemma 1. For any positive number $x>0$ there exists
  a positive integer $N$ such that $x>1/N>0$. 

Now, we want an $\varepsilon$ such that $c^2+5\varepsilon<2$, i.e., an $\varepsilon$ such that $\varepsilon<(2-c^2)/5$, which exists by Lemma 1.
In your case, since $0<(2-c^2)/(c+2)$, there is a $N$ such that $1/N<(2-c^2)/(c+2)$. Now, we can define $\varepsilon:=1/N$.

The key is, supposing that there is such $\varepsilon$, find some relation between $(x+\varepsilon)^2$ and $x^2+K\varepsilon$. Suppose $x^n<y$. There is many relations:


*

*First (Binomial formula): $$\sum_{k=0}^n\binom{n}{k}x^{n-k}\varepsilon^k.$$ Then use the expansion to obtain $(x+\varepsilon)^n\le x^n+x^{n-1}\varepsilon+\dotsb<y$.

*Second: $$(x+\varepsilon)^n\le x^n+\varepsilon((x+1)^n-x^n).$$
Then you can find $\varepsilon$ such that $(x+\varepsilon)^n<x$, just take a $\varepsilon$ such that $$0<\varepsilon<\min\left\{\frac{x-x^n}{(x+1)^n-x^n},1\right\},$$
which surely exists.

*Third: $$(x+\varepsilon)^n\le x^n+k\varepsilon$$
for some $k\in\Bbb R$. Then you obtain $(x+\varepsilon)^n\le x^n+k\varepsilon<y$, as desired.


All formulae are proved by induction on $m$.

A way to obtain this type of relations is see the behavior when $n$ increase.
For instance, let $x,y>0$ be rational numbers, and let $n\ge1$ be a integer number. We want to find some relation of the form $(x+\varepsilon)^n\le x^n+K\varepsilon$ for some $\varepsilon$ between $0$ and $1$.
Note that $\varepsilon^n\le\varepsilon$ for every $n$. Also, note that $\varepsilon$ always exists by Archimedean property.
Using some algebra, we can expand $(x+\varepsilon)^n$ when $n=1,2,3,4,5,\dots$
$$\begin{align}(x+\varepsilon)^2&=x^2+2x\varepsilon+\varepsilon^2\\&\le x^2+2x\varepsilon+\varepsilon\\&=x^2+\varepsilon(2x+1)\\\\(x+\varepsilon)^3&=x^3+3x^2\varepsilon+3x\varepsilon^2+\varepsilon^3\\&\le x^3+3x^2\varepsilon+3x\varepsilon+\varepsilon\\&=x^3+\varepsilon(3x^2+3x+1)\\\\(x+\varepsilon)^4&=x^4+4x^3\varepsilon+6x^2\varepsilon^2+4x\varepsilon^3+\varepsilon^4\\&\le x^4+4x^3\varepsilon+6x^2\varepsilon+4x\varepsilon+\varepsilon\\&=x^4+\varepsilon(4x^3+6x^2+4x+1)\\&\;\;\vdots\end{align}$$
So we have
$$\begin{align}(x+\varepsilon)^2&\le x^2+\varepsilon(2x+1)\\(x+\varepsilon)^3&\le x^3+\varepsilon(3x^2+3x+1)\\(x+\varepsilon)^4&\le x^4+\varepsilon(4x^3+6x^2+4x+1)\\(x+\varepsilon)^5&\le x^5+\varepsilon(5x^4+10x^3+10x^2+5x+1)\\(x+\varepsilon)^6&\le x^6+\varepsilon(6x^5+15x^4+20x^3+15x^2+6x+1)\\&\;\;\vdots\end{align}$$
Now, suppose that $x\le1$; so $x^n\le 1$ for every $n$. Thus
$$\begin{align}(x+\varepsilon)^2&\le x^2+3\varepsilon\\(x+\varepsilon)^3&\le x^3+7\varepsilon\\(x+\varepsilon)^4&\le x^4+15\varepsilon\\(x+\varepsilon)^5&\le x^4+31\varepsilon\\(x+\varepsilon)^6&\le x^4+63\varepsilon\\&\;\;\vdots\\\end{align}$$
Clearly, the relation from this is $$(x+\varepsilon)^n\le x^n+(2^n-1)\varepsilon.$$ (We can prove this by induction.)
Similarly, suppose $x>1$; so $x^n\ge x$ for every $n$. We can prove the relation $(x+\varepsilon)^n\le x^n+(2^n-1)x^{n-1}\varepsilon.$
But, if we don't want divide in cases, we can prove the relation $$(x+\varepsilon)^n\le x^n+n(2^n-1)(1+x)^n\varepsilon.$$
(To see this relation, try to prove the above relations and think how avoid the cases when some $k$ ensures the relation $(x+k)^n\ge x+k$.)
