Let $X_{n}$ be a sequence of real random variables and let $X$ a real random variable defined over the same probability space $(\Omega,\mathcal{A},\mathbb{P})$ and such that $X_{n}(\omega)\searrow_{n} X(\omega)$ for all $n\in\mathbb{N}$. Let $\{x_{n}\}_{n}$ be a sequence of real numbers such that $x_{n}\searrow_{n} x$ and such that $x_{n}$ is a median of $X_{n}$ for all $n\in\mathbb{N}$, that is,
$$\mathbb{P}(X_{n}\leq x_{n})\geq 1/2 \text{ and } \mathbb{P}(X_{n}\geq x_{n})\geq 1/2$$
The question is whether or not we have that
$$\mathbb{P}(X\geq x)\geq 1/2 - \epsilon,$$
for every $\epsilon > 0$.
$P(X<x) \leq \lim \inf P(X_n <x)$ (by convergence in distribution). Since $x \leq x_n$ this gives $P(X<x) \leq \lim \inf P(X_n <x_n)=\lim \inf [1-P(X_n \geq x_n)] \leq 1-\frac1 2$. Hence $P(X\geq x) =1-P(X<x) \geq \frac1 2$ which is stronger than what you want to prove.