Here is the question:
Let $s_n \rightarrow s$ and $c$ be any number strictly less than $s$. Prove that $s_n > c$ for all but finitely many $n$.
Attempt at a solution:
Since $s_n \rightarrow s$, we can say that $|s_n - s| \leq \epsilon$ for some $\epsilon > 0$. Let this $\epsilon = s - c$, which we know to be greater than $0$ since $s > c$. Now we have that $|s_n - s| \leq s - c$.
I tried applying the triangle difference inequality, but couldn't figure out how to work with the absolute values. Any help would be much appreciated!
As $s_n\to s$, take $\epsilon=s-c>0$. There exists a natural number $N$ such that for all $n$ such that $n\geq N$, we have $|s_n-s|\leq\epsilon$.
This means $-\epsilon \leq s_n-s\leq \epsilon$ so you get $c\leq s_n$ as desired by adding $s$ to each side.
The technique you missed is to replace the absolute value by its definition.
As suggested by a comment, you can also prove it by contradiction. Suppose there are infinitely many $n$ such that $s_n<c$. By picking those $n$ you can extract a subsequence of $(s_n)$, call it $(b_n)$, that is bounded by $c$ and still converges -- as a sub-sequence of a converging sequence -- to a certain value $b$.
As for all $n$ we have $c_n\leq c$ you have $b\leq c$ but by the uniqueness of the limit you also have $b=s>c$ which is a contradiction.