I've been working through a proof of this following theorem in Ross's Real Analysis textbook, and was hoping someone could look this over.
Theorem. If the sequence, $s_n$, converges, and if $s_n \geq a$ for all but finitely many $n$, then $\lim s_n \geq a$.
Proof. Let $\lim s_n = s$, and assume that $s_n \geq a$ for all but finitely many $n$. Then, assume further, to establish the logically equivalent contrapositive of this theorem, that $\lim s_n < a$. Since $\lim s_n = s$, we have that \begin{equation} \forall \epsilon >0, \exists N \in \mathbb{N}, \forall n > N, \left \lvert s_n - s \right \rvert < \epsilon. \end{equation} Let's choose $N$ so that $\left \lvert s_n - s \right \rvert < - s + a$. This implies that $s - a < s_n - s < - s + a$, and thus that $s_n < a$. Therefore, $\lim s_n \geq a$.
How does this look?
Consider the following proof by contradiction.
Let $s$ be the limit of $s_n$, to prove $s \geq a$, assume its contrary $s < a$. In this case, take $\varepsilon_0 = a - s > 0$ (fixed positive!), then for all $n \in \mathbb{N}$, it follows that $$|s_n - s| \geq s_n - s = s_n - a + a - s \geq 0 + \varepsilon_0 = \varepsilon_0,$$
which is clearly in contradiction to $\lim_n s_n = s$. This completes the proof.