Let $X$ be a topological space. If $X\times[0,1]$ is $T_{4}$ then the cone $CX$ is $T_{3\frac{1}{2}}$.
I tried to prove this using the definition of $T_{3\frac{1}{2}}$ but I really don't see how to manipulate the closed sets of the cone $CX$. Also, I was trying to understand the cone $CX$ like a decomposition space (i.e. like a partition of the space $X$, maybe like $\{\{\{(x,t)\}:x\in X,t\in[0,1)\},X\times\{1\}\}$ equipped with the quotient topology). Any help will be appreciated. Thank you.
P.D. This is an excercise of my last exam in my general topology course.
I have an idea how to see that. I would use Lemma of Urysohn to get $4_{\frac{1}{2}}$. Now of you want to separate a closed set from a point in the cone, you can look at the preimage of those two sets under the projection map. With $T_2$ you get that the point is closed in the cone so the preimages are closed sets, as the projection map is continuous. There you can separate the with a function (Urysohn). And then you find a unique continuous function from the cone into $[0,1]$ that commutes with your first one and the projection map. This one separates the point from the closed set. Is it clear what I meant?