Analysis Limit- Function Proof A) For every sequence ($p_n$) in $E$ such that $p_n \not= p$ and  $p_n \rightarrow p$ as $n \rightarrow \infty$ we have that
$f(p_n) \rightarrow l$ as $n \rightarrow \infty$ . ($E \subseteq \mathbb{R}$)
B) $f(x) \rightarrow l $ as $x \rightarrow p$
Prove that $A \Rightarrow B$

Proof:
Argue by contradiction. 
Suppose $f(x) \rightarrow l$ is not true. Then
$\exists \epsilon_0 > 0$, such that $ \forall \delta  > 0$,—which we choose to be $1/n$ for arbitrary $n$— $\exists x_n \in E$  ,
with $0 < |x_n − p| < 1/n$ but
$|f(x_n) − l| > \epsilon_0$.
Therefore we have found a sequence ($x_n$) which converges to $p$ but $f (x_n)$ does not
tend to $l$. Contradiction.

I mostly get this proof on the general level but it's the nitty gritty that troubles me. I have finished first year of a maths degree and I need to get into the habit of digesting every little detail of proofs. I will write in numbered list of questions and I will also try to justify why i think it might be the case. I would appreciate a reponse also in numbered format addressing each point, thanks.
1) Everything up to the second line is fine since it's just negating the definition of tending to a limit. 
But why- why choose $\delta$ to be $1/n$? 
Why do we have this freedom? 
Why $1/n$ specifically?
Is it because we know it's true $ \forall \delta $  and since a sequence is a map from the natural numbers we need a discrete sequence which matches each $x_n$ i.e   $x_1$ to $\frac{1}{1}$ and $x_2$ to $\frac{1}{2}$ etc.
Also because $1/n$ tends to $0$ any decreasing sequence would have sufficed?
2) Why does $x_n$ tend to $p$?
Since $1/n$ tends to $0$ it must be because of Simple Sandwiching or Squeeze Theorem right? 
So by Squeeze Theorem $|x_n-p| \rightarrow 0$ and so $x_n-p \rightarrow 0$ (Easy definition chase)
and so $x_n$ tend to $p$ (Easy definition chase)
 A: Yes; you answered your questions correctly. Since my post (up to here) would not be an answer material, I'll elaborate on $1)$.
We have that $\exists \ \epsilon_0$, such that for any chosen $\delta > 0$, there's an $x$ for which $0 < |x - p| < \delta$ and $|f(x) - l| \ge \epsilon_0$ hold (this is the crude form of the negation). Now we start to play.
Since this holds for any $\delta > 0$, choose $\delta = 1/1$ and call it $\delta_1$. Then, there's some $x$, call it $x_1$, such that $0 < |x_1 - p| < 1/1$ and $|f(x_1) - l| \ge \epsilon_0$. Do the same process, but with $\delta = 1/2$ and call it $\delta_2$, and name the corresponding $x$ as $x_2$. 
Continue doing this process. The axiom of choice tells us that we can go on forever doing this, and so obtaining the desired result. 
A: At its core, the proof is pointing out that sequences are enough to determine limits in the large.  
Let's talk about your proof then.  Suppose I didn't know what we were going to do.  As with the first line, you negate the definition and you have that $\exists \epsilon_0 > 0 $ so that, for any $\delta > 0$ there is some $x_{\delta} \in E$ with $|x_{\delta} - p| < \delta$ but $|f(x_n) - l| \ge \epsilon_0$.  
The whole "choose $\delta = \frac{1}{n}$ " bit is basically just the following.  We want to get a contradiction, so we'd like to contradict 1). So what we'd like to do is construct a sequence $x_n \to p$ for which we get $f(x_n) \not \to l$.
But we pretty much see we can do that at this point, because the above statement just says "no matter how close you get to $p$, you can find a point $x$ so that $f(x)$ is not near $l$".  So we'll choose such a point, say $x_1$ which is close to $p$ but with $f(x_1)$ at least $\epsilon_0$ away from $l$.  Are we done?  No, because maybe if we get closer to $p$ we'll be unable to avoid $l$.  But we know we can pick an even closer point $x_2$ so that this doesn't happen and so on.  The particulars of choosing points $x_n$ so that $0 < |x_n - p| < \frac{1}{n}$ is mostly just a formality.  We just need some sequence $\{x_n\}_n$ with $x_n \to p$ and $|f(x_n) - l| \ge \epsilon_0$.  Which points you pick is entirely up to you.
Why do we know that $x_n \to p$?  Because we're choosing them precisely to satisfy $0 < |x_n - p| < \frac{1}{n}$
