Let $E, F \subseteq \mathbb{R}$ such that $E \leq F$. Let $y \in \lim(F)$ be arbitrary and suppose to the contrary that $\lim(E) \leq y$ is not true. Then there is at least one $x \in \lim(E)$ such that $x \leq y$ is not true. Result R1008: Strictly ordered set of real numbers is trichotomic then states that $y < x$ must be true instead. That is, we have the inequalities $E \leq y \leq x$. If $\varepsilon > 0$, then $(y, x + \varepsilon)$ is now an open neighbourhood of $x$ which does not meet $E$. This contradicts the assumption that $x$ is a limit point of $E$. Therefore, we conclude $\lim(E) \leq y$. Since $y \in \lim(F)$ was arbitrary, we further conclude $\lim(E) \leq \lim(F)$. The second claim is a particular case of the first where $F$ is a singleton. $\square$