Example of a proper map $f: X\to Y$ such that $Y$ is normal, but $X$ is not The problem is:
Give an example of a proper map $f\colon X\to Y$ such that $Y$ is normal,
but $X$ is not.
My proposal is $f\colon\mathbb{R}_s\times\mathbb{R}_s\to\mathbb{R}_s$, where $\mathbb{R}_s$ is the Sorgenfrey line.
According to me this works, but I do not know how to prove it and I do not see a more concrete or easy example
 A: Projections can only be perfect if they're along a compact space:
$\pi_X: X \times Y \to X$ is perfect iff $Y$ is compact. Necessity  is clear as $\pi_X^{-1}[\{x\}] = \{x\} \times Y \simeq Y$ should be compact.
The Kuratowski theorem says that $\pi_X$ is a closed map for compact $Y$, which shows sufficiency, as closed maps with compact fibres preserve compactness under inverse images.
This idea can be used to create an example: I'll assume normal needs $T_1$ as well: otherwise there is a trivial example: 
Let $X$ be $\mathbb{N}$ in the cofinite topology and let $Y$ be $\mathbb{N}$ in the indiscrete (or trivial) topology. Then $Y$ is normal, because the only pair of disjoint closed sets is $\emptyset$ and $\mathbb{N}$, which are separated by themselves. All subsets of $Y$ are compact and the same holds in $X$, so just taking $f$ to be the identity we get a proper and continuous map, which is not closed. $X$ is not normal as we cannot separate $\{1\}$ and $\{2\}$ by disjoint open sets, e.g.
So using trivial spaces we can do it easily, as we saw. But if we up our standards we can find nicer examples, with the projection idea: 
Let $X = \omega_1$ (the set of countable ordinals in the order topology which Munkres calls $S_\Omega$, I believe) and $Y = \omega_1 + 1$ (or $\overline{S}_\Omega$). It is well-known that $Y$ is compact Hausdorff and $X$
is hereditarily normal (completely normal), but $X \times Y$ is not normal (we cannot separate the closed sets $\Delta = \{(\alpha, \alpha) : \alpha < \omega_1\}$ and $T = \{\alpha, \omega_1): \alpha < \omega_1\}$ by disjoint open sets.
By the above fact: $\pi_X$ is a proper map from the non-normal $X \times Y$ onto the hereditarily normal space $X$. (This product is example 2 in chapter 25 of Munkres (2nd ed.), as I saw later).
We can even find such spaces such that all fibres have size at most $2$ (idea from Engelking 3.12.20(e)): let $U \subset X \times Y$ be the (also non-normal) subspace $\{(\alpha, \beta): \alpha \le \beta < \omega_1\}$ and $f$ the canonical projection map defined on $U$ for the equivalence relation $R$ that only identifies for every $\alpha < \omega_1$ the points $(\alpha, \alpha)$ and $(\alpha ,\omega_1)$. $Y$ is the quotient under this relation in the quotient topology. Try to see it is normal, while $U$ in the subspace topology is not.
So if normal includes closed points, and you know about the $\omega_1 \times (\omega_1 + 1)$ being non-normal: use the projection onto the first factor.
The final example is just to see how badly things can go wrong.
If normal does not imply closed points, use the cofinite-indiscrete example, which is more elementary.
