Proofs in Real Analysis are too 'convenient'

These kinds of proofs rely on careful analysis of what you want to prove. Usually, you want to prove something of the form:

For every number $A$ there exists a number $B$ with certain properties related to $A$.

To prove such a statement, you take the number $A$ as a given because the theorem requires the statement to hold for all $A$. So no convenient choices allowed here. But then your task is to find one specific possible choice for $B$, because the statement is only that at least one such number exists.

For this reason it is often perfectly fine to make convenient choices in order to construct one specific convenient choice for $B$.

lets address the first proof...

There are two epsilons hiding in this proof.

One for the proposition we seek to prove $\forall \epsilon>0, \exists N \in \mathbb{N}$ such that $n>N\implies |x_n|<\epsilon$

We cannot choose the value of this epsilon.

But there is a second epsilon...

$\forall \epsilon>0, \exists N \in \mathbb{N}$ such that $n>N\implies \left||\frac {x_{n+1}}{x_n}| - l\right|<\epsilon$

This comes because it is given that this limit exists, and it is true for all values of $\epsilon.$ This means that we can choose the value for epsilon here and apply the results it generates to evaluate the other limit.

$\epsilon$ is a small number, not "any" number. Of course adding more words will help. The proof you posted assume that the reader has many background knowledge. For example:

Theorem: Let $\{x_n\}$ be a sequence of $\mathbb R^+$ such that $\lim_{n\to\infty} |\frac{x_n+1}{x_n}| = l$. If $0\le l \lt 1$, $\lim_{n\to\infty } x_n = 0$.
Proof: Consider $\epsilon \gt 0$ such that $l+\epsilon \lt 1$.
There exists $N \in \mathbb N$ such that $|\frac{x_n+1}{x_n}| \lt \epsilon$ for all $n \ge N$.

Here, "$\epsilon \gt 0$ such that $l+\epsilon \lt 1$" means that for any positive $\epsilon$ small enough that $l + \epsilon \lt 1$, the conclusion holds. We can always find at least one such $\epsilon > 0$ since $l < 1$ and so picking $0 < \epsilon = \frac{1 - l}{2}$ works.

I hope it helps!

Reply to the comments:

The logic go like this:

We want to prove that there exists $\epsilon$ such that $l+\epsilon<1$

To prove the existence, one example suffice. One example can be $\epsilon=\frac{1-l}{2}$.

In fact, for any $\epsilon<1-l$, this will also work.

You cannot arbitrarily choose $\epsilon$ from the whole real line in this example.

DougM's "choose" means "arbitrarily choose"; to further clarify, his "choose $\epsilon$" means that $\epsilon$ can be chosen from any positive real numbers.

In his second example the $\epsilon$ can be arbitrarily chosen from any positive numbers. In his first example, the $\epsilon$ can only be chosen from a restricted set of numbers.