Proof that Intertwined Sequence of two Convergent Sequences with different limits cannot be convergent Sorry for the mouthful of a title, I didn't know how else to describe the thing I'm trying to prove. I'm currently in the early stages of teaching myself Real Analysis with S. Abbott's book Understanding Analysis.
This question asks for a proof of the following conjecture:

Let $x_n$ be a convergent sequence with limit $x$. Let $y_n$ be a convergent sequence with limit $y$. Define a sequence $z_n$ that intertwines elements from $x_n$ and $y_n$. In other words, $z_n = {x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4...}$ Prove that $z_n$ is only convergent when $x = y$.

After thinking and writing a little bit, it becomes clear that this statement is intuitively obvious. As someone new to proofs and "Pure Mathematics" in general, I'm still having trouble deciding what I am allowed to assume is "obvious enough to not prove" and what I am not allowed to assume. I've seen certain proofs in the book where I see the author make certain intuitive assumptions, but sometimes those aren't allowed. It can be a little confusing sometimes.
My proof so far is making me question this yet again. My proof is as follows:
Proof:
Let $\epsilon_x$ define an arbitrarily small $\epsilon_x$-neighbourhood around $x$. Similarly, let $\epsilon_y$ define an arbitrarily small $\epsilon_y$-neighbourhood around $y$, where $\epsilon_x, \epsilon_y \in \mathbb{R}$.
If $x \ne y$, it is possible to choose $\epsilon_x$ and  $\epsilon_y$ such that the $\epsilon_x$-neighbourhood around $x$ and the $\epsilon_y$-neighbourhood around $y$ do not overlap.

By the definition of these neighbourhoods, it is that there exists some $N_x$ such that for any $n \ge N_x$, $x_n \in (-\epsilon_x, \epsilon_x).$ Similarily, there exists some $N_y$ such that for any $n \ge N_y$, $y_n \in (-\epsilon_y, \epsilon_y)$. In english, there exists some "points of no return" $N_x$ and $N_y$ in both $x_n$ and $y_n$ such that any point after must be within the arbitrary e-neighbourhood of the limit $x$ and $y$.
From this, it can be inferred that there is a point $N = max(N_x, N_y)$ such that if $n \ge N$, every point $z_n$ is either in the non-overlapping $\epsilon_x$-neighbourhood or the $\epsilon_y$-neighbourhood. In other words, there is a point $N$ such that all terms in the sequence $z_n$ must be in the x or y neighbourhoods, and not between or outside them.
For $z_n$ to converge to some point $z$, there needs to be, for an arbitrarily small $\epsilon_z$-neighbourhood around $z$, some point $N_z$ such that if $n \ge N_z, x_n \in (-\epsilon_z, \epsilon_z)$.
If $z$ is between $x$ and $y$, we can define an $\epsilon_z$-neighbourhood around $z$ that is small enough to fit between the two other x and y neighbourhoods. In this scenario, if $z_n$ indeed does converge to $z$, there must exist some point $N_z$ such that if $n \ge N_z, x_n \in (-\epsilon_z, \epsilon_z)$. But, as we stated previously, there already exists a point $N$ such that all points in $z_n, n \ge N$ are either in the $\epsilon_x$-neighbourhood or the $\epsilon_y$-neighbourhood, and not anywhere between. So, we cannot find some $N_z$ such that all points after $N_z$ are in the  $\epsilon_z$-neighbourhood, in other words, between the two neighbourhoods.

If $z$ is not between x and y, nearly the same argument can be made (all points after some point must be either in the x or y neighbourhoods).
So, if $x \ne y$, $z$ cannot converge.
On the other hand, if $x=y$, there exists no gap between two possible neighbourhoods that $z$ and its neighbourhood can be squished in. Therefore, in that case, it is possible to find a neighbourhood around $z$, where $z=x=y$, that all points after some point $N$ are located within.
First of all I am aware at how insanely ugly and even nonmathematical this proof is. I'm currently in high school so my exposure to proof techniques has been limited. I'm usually okay if provided with some direction, but I don't think I've had enough experience in more advanced mathematics to have the mathematical creativity to think of proofs without prompts.
I'm not even entirely sure this proof is valid, but after a few hours of working on this one problem its the one that has the least holes that I can see of.
Can anyone possible check the validity of this proof and offer a better one? I would be even more grateful to be tipped off on some of the problems with my proof format / etc...
Thanks!
 A: We'll prove the statement in both directions, i.e.

Let $(x_n)$ and $(y_n)$ be two given sequences and let $z_n$ be the sequence $(x_1,y_1,x_2,y_2,\dots,x_n,y_n,\dots).$ Then $(z_n)$ is convergent if and only if $(x_n)$ and $(y_n)$ are convergent with $\lim x_n=\lim y_n=L.$

$(\Rightarrow)$ Let $(z_n)$ be convergent with limit $L$. Then for every $\epsilon>0$, $\exists N\in \Bbb{N}$ such that for all $n\geq N$, $|z_n-L|<\epsilon.$ Notice that $x_n$ in $z_n$ are the odd index terms, while $y_n$ are the even index terms. Thus, it is certainly true that for our choice of $N$, $|z_{2n-1}-L|<\epsilon$ and $|z_{2n}-L|<\epsilon.$ Thus $\lim x_n=\lim y_n = L.$
$(\Leftarrow)$ Let $(x_n)$ and $(y_n)$ be convergent with common limit $L$. Then for every $\epsilon>0$, $\exists N_1$ such that $|x_n-L|<\epsilon$ for all $n\geq N_1$ and $\exists N_2$ such that $|y_n-L|<\epsilon$ for all $n\geq N_2$. Now pick $N=\max\{2N_1-1,2N_2\}$. This ensures that for this $N$, $|z_n-L|<\epsilon$ for all $n\geq N$.
A: Your proof seems valid to me and is what I initially thought to do, perhaps you need to show that it is possible to find the required neighbourhoods, but that can be done by making the radius less than half the distance between the closest "other" point among $x,y,z$.
Here is a possible proof by contrapositive. We show that if the intertwined sequence converges to $L$ then both sequences also converge to $L$.
Let $a_n$ be a sequence with $\lim\limits_{n\to \infty} a_n = L$, then for any subsequence $b_n$ we have $\lim\limits_{n\to \infty} b_n = L$ ( you would have to prove this but the proof is pretty clean).
Hence if the limit of the intertwined sequence is $L$ then both limits of each subsequence are also $L$.
