I know that even a non-Hausdorff first countable space is compactly generated, but I assume that adding the property that the space is also Hausdorff, there is an easier proof. How would you prove that a first countable Hausdorff space is compactly generated? I assume using the fact that a compact subspace in a Hausdorff space is closed is to key to make the proof easier, but I don't see how.
I use the following definition for a compactly generated space: A space is compactly generated if (i) a subspace $ A $ is closed in $ X $ if and only if (ii) $ A\cap C $ is closed in $ C $ for all compact subspaces $ C\subseteq X $.
To show that (i) $ \Rightarrow $ (ii) is easy. Since $ X $ is a Hausdorff space, $ C $ is closed and the intersection $ A\cap C $ is an intersection between two closed sets and hence closed in both $ C $ and $ X $.
What about the converse?