I will denote your space of all bounded real sequences by $X$ and let $M \subset X$ be as in your post. Also, for the sake of simplicity later, let's denote the $i$-th term of the sequence $z$ by $z(i)$ (instead of the usual $z_i$). Since you specifically asked about the norm, it is given by $\|z\|_X = \sup_{i \in \mathbb{N}} |z(i)|$
Let $x$ be the sequence defined by $x(n) = 1/n$ and let $x_m$ be the sequence defined by truncating the sequence $x$ after the first $m$ terms. That is, $x_m(n) = 1/n$ if $n \le m$ and $x_m(n) = 0$ otherwise.
Then we have that $x_m \in M$ for all $m$ and $\|x - x_m\|_X = \sup_{n \in \mathbb{N}}|x(n) - x_m(n)| = 1/(m+1)$
so that $\|x - x_m\|_X \to 0$ and hence $x_m \to x$ in $X$. So we have shown that $x_m$ is a convergent sequence in $X$. Therefore, it is also a Cauchy sequence. Moreover, the all the $x_m$ lie in $M$, but the limit $x$ is obviously not in $M$ (it is never $0$). Hence $x_m$ is a Cauchy sequence in $M$ which does not converge in $M$ (even though it converges in $X$).
In a sense, this is the only type of example you can come up with. Since $X$ is complete, a subset $M$ will be complete if it is closed which would mean, of course, that every Cauchy sequence in $M$ will converge in $M$. So you have to show that $M$ is not closed, which we did above by finding a convergent sequence of terms in $M$ whose limit is not in $M$.