Let $\displaystyle S$ be the set of all subsequential limits of $\displaystyle \left\{a_n\right\}$. Also let $\displaystyle \inf S=\sup S=\xi$. So it is clear that if $\displaystyle x\in S$ then $\displaystyle \xi\leqslant x \leqslant \xi\implies x=\xi$. Now let us define a subsequence of $\displaystyle \left\{a_n\right\}$ as $\displaystyle \left\{a_{\varphi}\right\}$ where $\displaystyle \varphi:\mathbb{N}\mapsto\mathbb{N}$ and $\displaystyle \varphi$ is strictly increasing. So since $\displaystyle \left\{a_{\varphi}\right\}\to\xi~\forall\varphi$ it is clear that there exists a $\displaystyle N$ such that $\displaystyle N\leqslant\varphi\implies d(a_{\varphi},\xi)<\varepsilon$.Now based on the above definition we may say that as soon as $\displaystyle \lceil\varphi^{-1}(N)\rceil\leqslant n\implies d\left(a_n,\xi\right)<\varepsilon$ which says exactly that not only does $\displaystyle a_n$ converge, but it converges to $\displaystyle \xi$.
That is how I would do it. Is that acceptable?
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