$(|a_n - b_n|) \rightarrow 0$ implies both sequences are either convergent or divergent 
Prove that if $(a_n)$ and $(b_n)$ are sequences such that $(|a_n -
 b_n|) \rightarrow 0$ then either: 1) both sequences are convergent to
  the same limit or 2) both sequences are divergent.

Is my following attempt at this proof correct?
The above statement can be logically rephrased as 
$$(|a_n - b_n|) \rightarrow 0 \implies \left[(a_n) \rightarrow c \iff (b_n) \rightarrow c\right]$$
Let $\epsilon > 0$. Now, $(|a_n - b_n|) \rightarrow 0$ implies that there exists $N_1 \in \mathbf{N}$ such that $|a_n - b_n| < \epsilon /2$ for all $n \ge N_1$. Now assume $(a_n) \rightarrow c$ which implies that there exists $N_2 \in \mathbf{N}$ such that $|a_n - c| <  \epsilon /2$ for all $n \ge N_2$. Set $N = \max\{N_1, N_2\}$, then whenever $n \ge N$ it follows that $|b_n - c| = |b_n - c -a_n+a_n| \le |a_n - b_n| + |a_n-c| < \epsilon/2 + \epsilon/2 = \epsilon$, hence $(b_n) \rightarrow c$. Now if we assume $(b_n) \rightarrow c$, then all we do is swap the $a_n$ and $b_n$ in the previous argument.

Your proof is entirely correct, and more than that, it's well written.
