$f$ proper but not universally closed

Let $D=\{0,1\}$ with the discrete topology, and let $K=\{0\}\cup\{2^{-n}:n\in\Bbb N\}$ with the topology that it inherits from $\Bbb R$. Let $X=K\times D$. Define an equivalence relation $\sim$ on $X$ by setting $\langle x,i\rangle\sim\langle y,j\rangle$ iff either $\langle x,i\rangle=\langle y,j\rangle$, or $x=y\ne 0$, and let $Y=X/\!\!\sim$. Let $q:X\to Y$ be the quotient map. Then $q$ is proper but not closed.

• $q$ is proper: If $C\subseteq Y$ is compact, then either $C$ is finite, in which case $q^{-1}[C]$ is finite, or $C\cap q[\{0\}\times D]\ne\varnothing$, in which case $\{0\}\times D\subseteq q^{-1}[C]$. In either case $q^{-1}[C]$ is compact.

• $q$ is not closed: Let $F=K\times\{0\}$. Then $F$ is closed in $X$, but $q(\langle 0,1\rangle)\in(\operatorname{cl}_Yq[F])\setminus q[F]$, so $q[F]$ is not closed in $Y$.

$Y$ is a simple sequence with two limits; it’s compact but not Hausdorff.