Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 2 / 4G (Metric and Topological Spaces)

True. In a discrete topology, the singletons form an open cover. If $X$ is compact, a finite subcover exists, so $X$ is finite. Conversely, any open cover of a finite set has a finite subcover: choose one member containing each point. This converse does not require the discrete topology.

False without a separation assumption. Give $X=\{0,1\}$ the indiscrete topology $\{\varnothing ,X\}$, and take $Y=\{0\}$. Both spaces are finite and hence compact, but $Y$ is not a closed set because $X\setminus Y=\{1\}$ is not open. In a Hausdorff space, a compact subset would be closed; that hypothesis is absent here.