Prob. 2, Sec. 1.4 in Kreyszig's Functional Analysis Book: Any Cauchy sequence with a convergent subsequence converges Here is Prob. 2, Sec. 1.4 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig. 

If $\left( x_n \right)$ is Cauchy and has a convergent subsequence, say, $x_{n_k} \to x$, show that $\left( x_n \right)$ is convergent with the limit $x$. 

My effort: 

Let $(X, d)$ be the metric space in which $\left( x_n \right)$ is a Cauchy sequence, and let $\varphi \colon \mathbb{N} \to \mathbb{N}$ be a strictly increasing function such that the sequence 
  $\left( x_{\varphi(n)} \right)$ converges to a point $x$ of $X$. And, we can put $$n_k = \varphi(k) \ \mbox{ for all } k \in \mathbb{N}.$$
Then, given a real number $\varepsilon > 0$, we can find a natural number $N$ such that $$ (1) \ \ \ \  d \left( x_m, x_n \right) < \frac{\varepsilon}{2} \ \mbox{ for any pair } (m, n) \mbox{ of natural numbers such that } m > N \mbox{ and } n > N.$$
Now for each $k \in \mathbb{N}$, we know that $n_k = \varphi(k) \in \mathbb{N}$; moreover, $$ n_k = \varphi(k) < \varphi(r) = n_r \ \mbox{ if $k \in \mathbb{N}$, $r \in \mathbb{N}$, and $k < r$},$$ because $\varphi$ is a strictly increasing function. 
Thus, $n_1 = \varphi(1) \in \mathbb{N}$ and so  $n_1 \geq 1$. Let $k$ be any natural number, and suppose that $$n_k = \varphi(k) \geq k.$$ Then since $\varphi$ is a strictly increasing function and since $k+1 \in \mathbb{N}$, therefore we can conclude that 
  $$n_{k+1} = \varphi(k+1) > \varphi(k) \geq k.$$
  But $\varphi(k+1) \in \mathbb{N}$. So we can conclude that $\varphi(k+1) \geq k+1$. Hence by induction it follows that $$n_k = \varphi(k) \geq k \mbox{ for all } k \in \mathbb{N}. $$
Thus we know that, for each $k \in \mathbb{N}$, (i) $\varphi(k) \in \mathbb{N}$ and (ii) $\varphi(k) \geq k$. 
Now since $$\lim_{k \to \infty} x_{\varphi(k)} = \lim_{k \to \infty} x_{n_k} = x, $$
  therefore we can find a natural number $K$ such that $$ (2) \ \ \ \ d\left( x_{n_k}, x \right) = d\left( x_{\varphi(k)}, x \right) < \frac{\varepsilon}{2} \ \mbox{ for any natural number } k \mbox{ such that } k > K.$$ 
Now let $M$ be any natural number such that $M > \max \left( K, N \right)$. Such an $M$ exists since the set of natural numbers is not bounded above (or by the Archimedian property of $\mathbb{R}$). 
Let $n \in \mathbb{N}$ such that $n > M$. Then we note that 
  $$n_{M+1} = \varphi(M+1) \geq M+1 > M > K$$ so that (2) can be used, and we also note that $$n_{M+1} \in \mathbb{N}, \ \ \ n \in \mathbb{N},  \ \ \ n_{M+1} > M > N, \ \ \ \mbox{ and } \ n > N$$
  so that (1) can be used. 
Then, from (1) and (2) above, we see that,
  $$ d\left( x_n, x\right) \leq d \left( x_n, x_{n_{M+1}} \right) + d \left( x_{n_{M+1}}, x \right) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon,$$
  for any natural number $n > M$.
Thus we have shown that, corresponding to  every real number $\varepsilon > 0$, we can find a natural number $M$ such that $$ d \left( x_n, x \right) < \varepsilon \ \mbox{ for any natural number } n > N.$$
  Hence the sequence  $\left(x_n \right)$ converges in the metric space $(X, d)$ to the point $x \in X$. 

Is the above proof correct? If so, then is the presentation good enough? If not, then where lies the flaw? 
 A: The above proof is correct, but there's a lot of parts that can be trimmed. For one, you don't really need to introduce the function $\phi$ at all. It just introduces confusion, and you can always write $n_k$ instead of $\phi(k)$ and lose no clarity.

So, a trimmed down proof:
Let $(x_n)_{n\in\mathbb N}$ be a Cauchy sequence, and let $(x_{n_k})_{k\in\mathbb N}$ be a convergent subsequence of it. Also, let $x=\lim_{k\to\infty} x_{n_k}$.
Let $\epsilon > 0$.
Then, 


*

*there exists some $N$ such that if $m,n\ge N$, we have $d(x_n, x_m)<\frac{\epsilon}{2}$.

*there exists some $K$ such that if $k\ge K$, we have $d(x, x_{n_k})<\frac\epsilon2$.


Set $N'$ to be the first value of the sequence $(n_K, n_{K+1},\dots)$ that is greater than $N$, and let $n>N'$. 


*

*From point (2), we know that $d(x, x_{N'})<\frac\epsilon 2$ 

*From point (1), we know that $d(x_{N'}, x_n)\leq \frac\epsilon 2$ (because $N'>N$ and $n>N$.


Then, $$d(x, x_n) \leq d(x, x_{N'}) + d(x_{N'}, x_n) < \frac\epsilon 2 + \frac\epsilon2=\epsilon$$ and the proof is over.
