Above is my question. My only issue is the final bit! For statements $1.$ and $2.$, the answer is true, since in both cases $Y$ is normal and we know that both metric and compact, Hausdorff spaces are normal. I'm assuming that $3.$ is false: firstly, it looks false (require normality for "standard" Urysohn, so unlikely to work for arbitrary topological space in the new case) and also the other two are true (unlikely to be three true!). However, I can't for the life of me come up with a counter example!
We know that for Urysohn if we define $$U = \{ x \in K \vert f(x) < 1/2 \} \, \ V = \{ x \in K \vert f(x) > 1/2 \},$$ then we obtain open, disjoint sets with $U \supseteq E$ and $V \supseteq F$ (where $E$ and $F$ are the closed, disjoint sets in Urysohn). This is for any such $E$ and $F$. So I thought that if I just define $X$ to be a non-normal space (eg, $X = \{ 0, 1, 2, ... \} = \Bbb N$), then I'd just be done since Urysohn fails... but in every case that I've tried while Urysohn fails, the extension property holds!
Any help would be most appreciated! As always, hints please, not just solutions!