About the structure of my $\varepsilon$-$\delta$ proof. I have been trying to study calculus and my textbook (namely Courant’s “Differential and Integral Calculus”) provides no examples of how to do an $\varepsilon$-$\delta$ proof whatsoever, though in the problem set it does ask me to perform some. I have written one down, and I know it is not perfect (maybe not even good or acceptable). So the main question is: $$ \textbf{What am I doing right or wrong, what I can improve in order for my proofs to be just?}$$. $$\textit{Prove:} \;\; \lim_{x \to \xi} (x^2 - 2x) = {\xi}^2 - 2\xi $$
$$\textit{Proof:} \;\; | f(x) - f(\xi) | = | x^2 - 2x - {\xi}^2 + 2\xi | = |(x^2 - {\xi}^2) - 2(x - \xi) | = | x - \xi | | x + \xi - 2 | = | x - \xi | | (x - \xi) + 2(\xi - 1) | \leq | x - \xi | (| x - \xi | + 2| \xi -1 |) $$
Thus, if $\forall \varepsilon > 0$ we find $\delta = \delta (\varepsilon)$ such that $$ | x - \xi | (| x - \xi | + 2| \xi -1 |) < \varepsilon \;\; \text{as} \;\; 0 < | x - \xi | <  \delta $$ , the proof shall be complete.
Hence, we can obtain that (by expanding the brackets and solving the quadratic inequality) $$ 0 < | x - \xi | < \sqrt{(\xi - 1)^2 + \varepsilon} - | \xi - 1 |^2 $$.
If we should put the right-hand side (which, as can be easily observed, is greater than 0 $\forall \varepsilon > 0$) of the obtained inequality to be equal to $\delta > 0$, then our proof will be complete.
All your help is appreciated, thank you!
P.S. I am extremely sorry for all the inconvenience caused by my lack of knowledge of LaTeX, I am doing by best!

Note : This should be a comment but it turns out I do not have enough reputation points so I cannot leave comments yet.
First of all, your right-hand side is not always positive : take $\epsilon=1$, $\xi=3$.
One alternative way of doing it is the following : find a positive $\delta$ such that both $\delta^2$ and $2\delta|\xi-1|$ are less than $\epsilon/2$. This is easily done by taking $\delta \lt \min(\sqrt{\epsilon/2},\frac{\epsilon}{4|\xi-1|})$.
This method is very efficient in general : if you can manipulate your expression so that you need to bound a sum of $n$ terms by epsilon, you only need to bound each term by $\epsilon/n$ simultaneously using the minimum trick.
