limit of diverging sequence proof I'd really appreciate it if you read my proof. It seems deceptively simple, I feel like I'm missing something

Two sequences $\{X_n\}$, $\{Y_n\}$ in a metric space satisfy $\lim_{n\to\infty} d(X_n,Y_n) = 0$. If $\{X_n\}$ diverges, then show that $\{Y_n\}$ also diverges.

If the distance between the limits of the two sequences is $0$, that means they must approach the same element:
$$
\lim_{n\to\infty} X_n = x \iff \lim_{n\to\infty} d(X_n,x) = 0
$$
If $\{X_n\}$ diverges, that means as $n$ approaches infinity, its limit is either $-\infty$ or $+\infty$, which will be the same limit as $\{Y_n\}$. Therefore, $\{Y_n\}$ diverges.
 A: You're almost there, but going wrong when thinking to “diverge” as meaning “the limit is infinity”. In a general metric space there's no “infinity” like in the reals.
I so assume that “diverging” is taken as a synonym for “not converging”.
Suppose $\{Y_n\}$ converges to $y$. Then you have, for every $n$,
$$
0\le d(X_n,y)\le d(X_n,Y_n)+d(Y_n,y)
$$
by the triangle inequality. So, by the squeeze theorem,
$$
0\le\lim_{n\to\infty}d(X_n,y)\le 
\lim_{n\to\infty}(d(X_n,Y_n)+d(Y_n,y))=0
$$
and therefore $\{X_n\}$ converges to $y$.
Similarly, if $\{X_n\}$ converges to $x$, also $\{Y_n\}$ converges to $x$.
Thus, by contrapositive, if one sequence does not converge, so does the other one.
A: I would go about proving the contrapositive. Assume $(y_n)$ converges to a point $p$. Then $\lim d(x_n,y_n)$ as $n\to\infty$ becomes $\lim d(x_n,p)$ as $n\to\infty$ which we know evaluates to $0$. Since it's a metric space, this only occurs when $x_n=p$. Therefore the $\lim d(x_n,p)$ as $n\to\infty$ becomes $\lim d(p,p)$. Therefore $(x_n)$ must converge to $p$ as well.
