Arbitrary intersection of closed compact sets is compact
We've been trying to find a counter example to this, however we failed. So we would be happy if someone can tell us if this proposition is correct or false, so we can stop wasting our time trying to find a counter example
Let $(Y_i)$ be a family of closed compact subsets of a space $X$. The intersection of closed sets is closed. Better yet, $\bigcap_i Y_i$ is closed in $Y_j$ for each $j$ by the usual argument for showing that a set is closed with respect to the subspace topology. Since a closed subspace of compact space is compact, $\cap_i Y_i$ is compact.