Prove that if $s_n ≤ b$ for all but finitely many $n$, then $\lim s_n ≤ b$. The question asks me to prove  that if $s_n ≤ b$ for all but finitely many $n$, then $\lim s_n ≤ b$ where $(s_n)$ be a sequence that converges.
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Here is how I did it but im not sure if its entirely correct. I used proof by contradiction.
Suppose $\lim s_n>b$ and $s_n \leq b$ for all but finitely many $n$. Let $S=\lim s_n$. By definition we have for every $n>N$ which implies $|s_n-s|<\epsilon$. Let $\epsilon= b-2s+s_n$.
Now we get $s_n-s<\epsilon=b-2s+s_n$ which simplifies to $s<b=\lim s_n<b$ which is a contradiction. 

Your idea of proof by contradiction might be better modified as follows:
If the limit of $s_n$ is greater than $b$, it can be written as $b+c$ for some $c \in \mathbb{R}^{+}$.
Since this is the limit of the sequence, we can find infinitely many $s_n$ that are arbitrarily close to $b+c$; in particular, we can find infinitely many $s_n$ that are greater than $b$, which contradicts the given that there are only finitely many such $s_n$.
Therefore, our supposition that the limit is greater than $b$ must have been incorrect, hence the limit is less than or equal to $b$ as desired. QED
