Let $u = \frac{a}{b}$ and $x = \frac{c}{d}$ be the reduced fractions of $u$ and $x$. We are looking for rational numbers $v$ such that $v = \frac{mc-a}{md-b}$, where $m$ is large enough integer such that $mc-a$ and $md-b$ are both positive. According to the definition, $x$ is the mediant of $u$ and $v$ as soon as $\frac{mc-a}{md-b}$ is irreducible. Let's show that there are infinitely many natural numbers $m$ such that $\frac{mc-a}{md-b}$ is irreducible. This will complete the solution.

Let us first prove a lemma: prime numbers which can be used to reduce the fractions are also the factors of $ad - bc$. Indeed, if $p \mid mc - a$ and $p \mid md - b$, then $p \mid a md - b - b (mc - a) = m(ad - bc)$, therefore $p \mid m$ or $p \mid ad - bc$. If $p \mid m$, then $p \mid a$ and $p \mid b$ which contradicts the irreducibility of the fraction $\frac{a}{b}$. Therefore $p \mid ad - bc$.

As $u$ and $x$ are different, $ad - bc \ne 0$. Therefore the number $ad - bc$ has a finite number of prime factors. Let $p_1, \dots, p_l$ be all the different prime factors which can reduce the fraction $\frac{mc-a}{md-b}$ for at least one $m$ and for each $i = 1, \dots, l$ let $m_i$ be natural number such that the fraction $\frac{m_i c - a}{m_i d - b}$ is reducible with prime $p_i$.

Let $n$ be an arbitrary factor for which the fraction $\frac{nc-a}{nd-b}$ is not irreducible. This fraction must be reducible with some prime number $p_i$ which can also reduce the fraction $\frac{m_i c - a}{m_i d - b}$. Then $p_i \mid (n - m_i)c$ and $p_i \mid (n - m_i)d$. Therefore $p_i \mid n - m_i$ as otherwise $p \mid c$ and $p \mid d$ which contradicts the irreducibility of the fraction $\frac{c}{d}$. Therefore $n \equiv m_i \pmod{p_i}$.

Therefore by choosing $n$ such that $n \equiv m_i + 1 \pmod{p_i}$ for each $i = 1, \dots, l$ the fraction $\frac{nc-a}{nd-b}$ must be irreducible. According to Chinese remainder theorem there are infinitely many natural numbers $n$ which satisfy such congruence system.