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I came across the following result in a book:

Let $(u_n)$ be a Cauchy sequence in the normed space $X$ over $\mathbb{K}$, which has a convergent subsequence $(u_{n_k})$, that is, $u_{n_k} \to u$. Then the entire sequence converges to $u$.

The proof given there is as follows:

Let $\epsilon>0$ be given. There is an $n_0(\epsilon)$ such that $$\|u_n-u_m\|<\epsilon~~\forall~n,m\geq n_0(\epsilon)$$ Since $(u_{n_k})$ converges to $u$, there exists some fixed index $m$ such that $$\|u_m-u\|<\epsilon,~~\text{where}~m\geq n_0(\epsilon)$$ By the triangle inequality, $$\|u_n-u\|\leq \|u_n-u_m\|+\|u_m-u\|\leq 2\epsilon~~\forall~n\geq n_0(\epsilon)$$ Hence $u_n \to u$ as $n \to \infty$.

I know that there's a similar question here, but the answer given there is more of a hint. The proof above isn't clear to me and seems to have glossed over some details that a beginner like me can't understand. I'm trying to write a complete proof and would appreciate if it could be verified:

Let $\epsilon>0$. Since $(u_n)$ is Cauchy, there's an $N$ such that $$\|u_n-u_m\|<\epsilon/2~~\forall~n,m\geq N$$

Since $(u_{n_k})$ converges, there's a $K$ such that $$\|u_{n_k}-u\|<\epsilon/2~~\forall~k\geq K$$

If $n_K \geq N$, then $\|u_n-u_{n_k}\|<\epsilon/2~~\forall~n\geq N$ and $\forall~k\geq K$. This means $$\|u_n-u\|\leq \|u_{n_k}-u\|+\|u_n-u_{n_k}\|<\epsilon~~\forall~n\geq N~~\text{and}~~k\geq K$$

If $n_K < N$, then we can always choose a sufficiently large $K'>K$ such that $n_{K'} \geq N$ and proceed as above, since we'd still have the $\|u_{n_k}-u\|<\epsilon/2$ inequality for all $k \geq K' > K$.

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    $\begingroup$ Your proof works perfectly well but could be made a bit more efficient. For example, instead of splitting into the cases $n_K < N$ and $n_K \geq N$, you could pick a $K \geq N$ straight away (since you don't need to choose the smallest such $K$). This is essentially what is done in the proof in the book. $\endgroup$ – Rhys Steele Jul 3 '18 at 11:07
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    $\begingroup$ Loose the if statements by $M=\max\{K,N\}$ since $d(u_n,u_m)<\epsilon/2$ for alle $n,m\ge M\ge N$ and $d(u_{n_k},u)<\epsilon/2$ for alle $k\ge M\ge K$. More precise. $\endgroup$ – bubububub Jul 3 '18 at 11:08

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