Let $x_n$ a sequence in a Banach Space $B$ such that give a $\epsilon>0$, there is a convergent sequence $\{y_n\}$ with $\|y_n-x_n\|<\epsilon$

Your attempt isn't correct. You have $\|x_n - y\| \le \|x_n - y_n\| + \|y_n - y\|$ so the best you can say is that $\limsup \|x_n - y\| \le \epsilon$.

You need to vary $\epsilon$. For each positive integer $k$ there is a sequence ${y_n^k}$ with limit $y^k$ satisfying $\|x_n - y_n^k\| < \frac 1k$ for all $n$. As in the first paragraph you have $\limsup_n \|x_n - y^k\| \le \frac 1k$. For any two indices $j$ and $k$ you have $$\|y^j - y^k\| \le \|x_n - y^j\| + \|x_n - y^k\|$$ for every index $n$, and by calculating the limsup with respect to $n$ it follows that $$\|y^j - y^k\| \le \frac 1j + \frac 1k.$$ This means that the sequence of limits is Cauchy. Since this is a Banach space it has a limit $y$.

Finally examine $\|x_n - y\|$. For any index $k$ you have $$\|x_n - y\| \le \|x_n - y^k\| + \|y^k - y\|$$ so that $$\limsup_n \|x_n - y\| \le \limsup_n \|x_n - y^k\| + \|y^k - y\| \le \frac 1k + \|y^k - y\|.$$ Let $k \to \infty$ to discover $\limsup_n \|x_n - y\| \le 0$.