Prove that $(1+\frac{1}{n})^n$ is a Cauchy sequence

If all you want to do is prove that $\mathbb{Q}$ is not a complete metric space, then there are much easier sequences you can utilize than $\left(1+\frac{1}{n}\right)^n$. For example, consider the sequence

$$a_n=\frac{\lfloor 10^n \sqrt{2}\rfloor}{10^n}$$

This sequence converges to $\sqrt{2}$ since

$$\sqrt{2}-\frac{1}{10^n}=\frac{10^n\sqrt{2}-1}{10^n}\leq a_n\leq \frac{10^n\sqrt{2}}{10^n}=\sqrt{2}$$

but we also have $a_n\in\mathbb{Q}$ for all $n$. Since $\sqrt{2}$ is irrational we are done.

EDIT: Here is a proof that a sequence is convergent if and only if its Cauchy.

Suppose that $a_n\to L$ and let $\epsilon>0$ be given. By definition, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $|a_n-L|<\frac{\epsilon}{2}$. But then for $n,m\geq N$ we have

$$|a_n-a_m|=|a_n-L-a_m+L|\leq |a_n-L|+|a_m-L|<\epsilon$$

Now, suppose that $a_n$ is Cauchy and let $\epsilon>0$ be given.

Part $1$: Convergent subsequence

From the definition, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $|a_N-a_n|<1$. This then implies that for all $n\geq N$ we have

$$a_N-1<a_n<a_N+1$$

and therefore for all $n\in\mathbb{N}$

$$\min\{a_1,a_2,...,a_{N-1},a_N-1\}\leq a_n \leq \max\{a_1,a_2,...,a_{N-1},a_N+1\}$$

This implies that $a_n$ is a bounded sequence. Then by the Bolzano–Weierstrass theorem we may conclude that $a_n$ has at least one convergent subsequence. Let us denote this subsequence by $a_{n_k}$ and let its limit be

$$\lim_{k\to\infty} a_{n_k}=L$$

Part $2$: Entire sequence converges

Now, let $\epsilon>0$ be given. By definition, there exists $N_1\in\mathbb{N}$ such that $k\geq N$ implies $|a_{n_k}-L|<\frac{\epsilon}{2}$. Additionally, there exists $N_2\in\mathbb{N}$ such that $N_2\leq n\leq m$ implies $|a_n-a_m|<\frac{\epsilon}{2}$. Now, choose $N=\max\{N_1,N_2\}$ and let $l$ be the smallest natural such that $n_l\geq N$ and $a_{n_l}$ is a member of the convergent subsequence. Then for all $n\geq N$ we have

$$\bigg||a_n-L|-|a_{n_l}-L|\bigg|\leq |(a_n-L)-(a_{n_l}-L)|=|a_n-a_{n_l}|<\frac{\epsilon}{2}$$

(the first step in this equation is the reverse triangle inequality). This then implies

$$0\leq |a_n-L|<|a_n-L|-|a_{n_l}-L|+\frac{\epsilon}{2}\leq \bigg||a_n-L|-|a_{n_l}-L|\bigg|+\frac{\epsilon}{2}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

(since $|a_{n_l}-L|<\frac{\epsilon}{2}$). We conclude that

$$\lim_{n\to\infty}a_n=L$$