Let $B$ be the set of all bounded real sequences. Prove that $(B,\rho)$, where $\rho(x,y):=\sup_{n}\dfrac{|x_n-y_n|}{n}$ is a non complete metric space.
I could show that this is a metric space and I also know that with the metric $d(x,y):=\sup_{n}|x_n-y_n|$ this space is complete. How do I prove that it is not complete with $\rho$?
The intuition you should have here is that the metric $\rho$ reduces the information you have about the tails of the sequences you're looking at so we should try to make something bad happen there.
For $k \leq n$, let $x_k^{(n)} = k^{1/2}$ and otherwise let $x_k^{(n)} = 0$. So then, for fixed $n$, $x^{(n)} \in B$. It is straightforward to show that $(x^{(n)})_{n \geq 1}$ is a Cauchy sequence for $\rho$ (in fact, this is why I chose something like $k^{1/2}$). I claim that $x^{(n)}$ does not converge in $(B, \rho)$ as $n \to \infty$.
To see this, first notice that if $\rho(x^{(n)}, x) \to 0$ then in particular, for each fixed $k$, $x_k^{(n)} \to x_k$. As a result, if $x^{(n)} \to x$ in $(B, \rho)$ then we must have $x = (1, 2^{1/2}, 3^{1/2}, 4^{1/2}, \dots)$. But this is not a bounded sequence and hence doesn't lie in $B$, giving us a contradiction.