Suppose the space is compact. Take any net $x_\alpha=f(\alpha)$ . For each $\alpha$ let $B_\alpha=\{\beta|\alpha\preceq\beta\}$ . Then, given that the index set is directed, the collection of sets $f(B_\alpha)$ satisfies the finite intersection property. Let $x$ be the intersection of the closures of these sets and $U$ be its neighborhood. Since $x$ lies in the closure of each $f(B_\alpha)$ , for each $\alpha$ there is $\beta\in B_\alpha$ such that $x_\beta\in U$ . Therefore, $x$ is an accumulation point of the net, and, using the previous exercise, there is a subnet converging to $x$ .The other direction. Suppose that every net in $X$ has a convergent subnet. Let $B_j$ , $j\in J$ be a collection of closed sets satisfying the finite intersection property. Consider the set $K$ of all finite subsets of $J$ : $K=\{\alpha\subseteq J||\alpha|<\infty\}$ . It is a directed set with the partial order given by the reverse inclusion. Moreover, since each finite intersection has a point, for each $\alpha\in K$ we can take $x_\alpha\in\cap_{j\in\alpha}B_j$ . This is a net in $X$ , and it must have an accumulation point $x$ . In particular, for every $k\in J$ and neighborhood $U$ of $x$ there must be $\beta\succeq\{k\}$ such that $x_\beta\in U$ . Given that $x_\beta\in\cap_{j\in\beta}B_j$ and $\beta\subseteq\{k\}$ we conclude that $U\cap B_k\neq\emptyset$ . Thus, $x$ lies within the closure of $B_k$ which is equal to $B_k$ . Therefore, $\cap_j B_j$ is not empty.