I tried to prove this but my proof did not match with my book's. So I want to verify whether my proof is correct or not.
Theorem: Every Cauchy sequence in $\mathbb R$ has a limit.
Let us assume the contrary that there is a sequence $(a_n)$ which is Cauchy but not convergent.
1.Since the sequence is not convergent,for all real $a$, there must be an $\epsilon$ such that for all $n \in \mathbb N$, $\exists n_0 \geq n$ such that $|a_{n_0}-a|\geq \epsilon$.
2.Since $(a_n)$ is Cauchy, we can show that that particular $\epsilon$ we talked above,there is a natural $N$ such that for all $n,m\geq N$, $|a_n-a_m|<\epsilon$.
3.Go and look $(1)$. I can thus find $m_0 \geq N$ such that $|a_{m_0}-a| \geq \epsilon >|a_{m_0}-a_n|$ for all $n \geq N$. Since $a$ is arbitrary, putting $a=a_n$,we get contradiction.Thus, the proof.
Best Answer
In step 1, you're fixing $n_0$ which has the property $|a_{n_0} - a| \geq \epsilon$, but in step 3, you're using that $|a_{m_0} - a| \geq \epsilon$ for your $m$ large enough.
You can't first fix $n_0$ and then replace it. That's already assuming that your sequence $a_n$ converges to $a$, which is what you're trying to prove.
The standard argument constructs (using Bolzano-Weierstrass) a convergent subsequence of $a_n$. It can then be shown that any Cauchy sequence with a convergent subsequence is convergent.
A quicker way to see that your argument doesn't work is the following: it doesn't rely on the "completeness" of $\mathbb{R}$. You will learn that - in more general settings - convergent sequences are always Cauchy, but not all Cauchy sequences converge. For this you really need the completeness (which is used in the proof of Bolzano Weierstrass).