Let $\omega \in \Omega$ be an outcome such that $\sum_{n = 1}^N X_n(\omega) \neq \sum_{n = 1}^N Y_n(\omega)$. Then $X_n(\omega) \neq Y_n(\omega)$ for at least one $n \in 1, \ldots, N$, for otherwise the sums $\sum_{n = 1}^N X_n(\omega)$ and $\sum_{n = 1}^N Y_n(\omega)$ would be equal. Thus, by definition of a union, $\omega \in \bigcup_{n = 1}^N \{ X_n \neq Y_n \}$. Since $\omega \in \Omega$ was arbitrary, we have the inclusion $$\begin{equation} \left\{ \sum_{n = 1}^N X_n \neq \sum_{n = 1}^N Y_n \right\} \subseteq \bigcup_{n = 1}^N \{ X_n \neq Y_n \} \end{equation}$$ Result R2090: Isotonicity of probability measure now implies $$\begin{equation} \mathbb{P} \left( \sum_{n = 1}^N X_n \neq \sum_{n = 1}^N Y_n \right) \leq \mathbb{P} \left( \bigcup_{n = 1}^N \{ X_n \neq Y_n \} \right) \end{equation}$$ which is what was required to be shown. $\square$