We have that $X$ is a set of all sequences $\{x_n\}_{n \in \mathbb{N}}$ of real numbers such that $\lim_{n \to \infty} x_n = 0$. Now we introduce $Y$ as a set of all sequences $\{y_n\}_{n \in \mathbb{N}}$ of real numbers such that $\sum_{n=1}^{\infty} |y_n| < \infty$.
I have to show that $Y \subseteq X$ and need to find a sequence $\{x_n\}$ that belongs only to $X$ and not $Y$.
To show that $Y \subseteq X$, all elements of $Y$ should be contained in $X$. I know that the sum of the absolute value of the sequences is smaller than $\infty$ which means that it's converging. Is this where I need to use Bolzano-Weierstrass Theorem to show that $Y \subseteq X$.
For the second part of the problem, would the sequence that only belongs to $X$ and not $Y$ be the supremum of $X$.
For the first part, the inequality
$$ |y_k| \leq \sum_{n=k}^{\infty} |y_n| $$
is useful. For the second part, the sequence $x_n = \frac{1}{n}$ is always a good place to start looking.