I was working on a problem where I should demonstrate that not every Cauchy seq. is convergent. So I choosed $(X,d) = (\mathbb{Q}, |\cdot|)$. Thus $x_n = (1+ 1/n)^n$ which we know lies in $\mathbb{Q}$ but it converges to $e \not \in \mathbb{Q}$. But still I have to prove that it is Cauchy in $\mathbb{Q}$.
So $|(1+1/n)^n - (1+1/m)^m|< \epsilon$ but I dont see how I can do that.
$(x_n)$ is a Cauchy sequence in $ \mathbb R$.
$x_n \in \mathbb Q$ for all $n$.
It follows: $(x_n)$ is a Cauchy sequence in $ \mathbb Q$.
If you insist on this direct approach, it is possible. However, there will be a lot of wading through coefficients using the binomial expansion. For a sketch version, remember that we add fractions by finding a common denominator:
\begin{align*} \left(1+\frac{1}{n} \right)^n - \left(1 + \frac{1}{m} \right)^m &= \left(\frac{n+1}{n} \right)^n - \left( \frac{m+1}{m} \right)^m \\ &= \frac{(n+1)^n}{n^n} - \frac{(m+1)^m}{m^m} \\ &= \frac{m^m (n+1)^n - n^n (m+1)^n}{n^n m^m} \\ &= \frac{(m^m n^n + \cdots + m^m \cdot 1) - (n^n m^m + \cdots +n^n \cdot 1)}{n^n m^m} \end{align*} There will be a lot of leftover terms after one easy subtraction, involving binomial coefficients. You will need to show that these remaining terms are all "small" compared with the $n^n m^m$ term in the denominator.
Alternative
Since you settled on $e \not \in \mathbb{Q}$, how about the following sequence: $$a_0 = 2, ~a_1 = 2 + \frac{7}{10}, ~a_2 = 2 + \frac{7}{10} + \frac{1}{10^2}, ~a_3 = 2 + \frac{7}{10} + \frac{1}{10^2} + \frac{8}{10^3}, \dots $$
Without loss of generality, take $n>m$. Letting $d_i$ be the $i^{th}$ digit in the decimal expansion of $e$, $$a_n - a_m = \frac{d_{m+1}}{10^{m+1} } + \cdots + \frac{d_n}{10^n}. $$ Since $d_i \leq 9$ for all $i$, we can explicitly bound this by using a geometric series. This will be much easier.
The Kicker
Either way, you have a Cauchy Sequence in $\mathbb{Q}$ converging to some number. Since you're insisting doing this from first principles, you're only halfway done. How do you know that $e$ is not rational? Have you ever personally proven this? To give a hint that there's some difficulty, consider the following timeline:
Ancient Greeks had proven that $\sqrt{2}$ is irrational
Jacob Bernoulli introduces $e$ in 1683
Euler proves $e$ is irrational in 1737 (50 years later)
Lambert proves $\pi$ is irrational in 1761
At a first glance, $e= 2.718281828\dots$ seems like a repeating decimal, no? But aren't repeating decimals rational numbers? =P
