This is a problem that becomes much simpler if one makes the effort to make the problem precise. The following is somewhat inspired by [Alexander Gnedin. "Guess the larger number." arXiv:1608.01899 (2016).] that contains a lot of discussion, variations, and history regarding the problem.
A pure strategy specifies for each observed number whether one accepts it as the larger or at least as large one ($1$) or passes to the other number ($0$). So we can take the space of pure strategies to be $\{0,1\}^\mathbb{Z}$. For each let $t_n\in\{0,1\}^\mathbb{Z}$ be the $n^\text{th}$threshold strategy for which $t_n(l)=1$ iff $l\geq n$.
Now let $\nu$ be any probability distribution over $\mathbb{Z}$ and let $\pi(\nu,t)$be the winning probability of picking the larger or equally larger number of someone using the pure strategy $t$ when both numbers are chosen independently from $\nu$. (Note that this does not fully agree with the original statement in which both numbers must be distinct, but if we have enough freedom to pick both distributions separately, one cannot ensure a winning probability of more than $1/2$ for each $\nu$.)
It is easy to see that for each $\nu$, the winning probability $\pi$ is maximized at some threshold strategy and that any other maximizing strategy must equal a threshold strategy $\nu$-almost surely.
Observe that a threshold strategy guarantees a winning probability of at least $1/2$ against any distribution $\nu$ on the integers. Also, for any distribution $\nu$ on the integers, some threshold strategy has a winning probability strictly larger than $1/2$.
So far, nothing should have been truly surprising. Now let $\mu$ be any full support probability distribution on $\mathbb{Z}$. There is an induced randomized threshold strategy$t_\mu$ that chooses the threshold strategy $t_n$ with probability $\mu(n)$. If we extend $\pi$ in the obvious way to a function $\pi:\Delta(\mathbb{Z})\times\Delta(\{0,1\}^\mathbb{Z})\to\mathbb{R}$ where $\Delta(X)$ is the set of Borel probability measures on $X$, we see that $\pi(\nu,t_\mu)>1/2$ for all $\nu\in\Delta(\mathbb{Z})$. Indeed, for each $\nu$, and each $n$, $\pi(\nu,t_n)\geq 1/2$, and with positive probability, we pick some $n$ such that $\pi(\nu,t_n)>1/2$.
Finally, note that the originally proposed strategy based on a continuously distributed random variable $r$ corresponds to $t_\mu$ with $\mu(n)$ being the probability that $n>r$.