The space $X$ below is not compact. Find explicitly a sequence of closed sets $(F_n)_{n=1}^{\infty}$ with the finite intersection property, but such that $\bigcap_{n=1}^{\infty}F_n=\emptyset$.
a) $X = \{0\} \cup (1/n)_{n=1}^{\infty}$ with the metric $d(x,y)=|x-1|+|y-1|$ for $x \neq y, \enspace d(x,x)=0$.
b) $X = \{(x_n)_{n=1}^{\infty}: \sum_{n=1}^{\infty}|x_n|\leq 1 \}$ with $d(x,y)=\sum_{n=1}^{\infty}|x_n-y_n|$.
Here is my attempt.
a) We can construct $F_n=\{1/3+1/k: k > n\}$ such that $F_{n+1} \subset F_n$ for $\forall n \in \mathbb{N}$. This is the nested interval of closed balls and it is clear that $\bigcap_{n=1}^{\infty} F_n = \emptyset$ by Cantor's intersection theorem, because we do not have the condition that $r_n \rightarrow 0$ as $n \rightarrow \infty$.
b) We can construct $F_n=\{x \in X : x_k=0$ for $k \leq n-1, \sum_{n=1}^{\infty}|x_k| \geq1 \}$. Therefore, $F_{n+1} \subset F_n$ for $\forall n \in \mathbb{N}$ and $\bigcap_{n=1}^{\infty} F_n = \emptyset$.
Is my approach correct? There is another thing that bothers me with part b. The set $X$ is bounded and closed. But how come isn't it compact?