I am trying to find the error in my proof. I have used two separate cases of the triangle inequality to determine that $|s_n - s_m| < \epsilon \implies |s_n - L|< \epsilon$, for some $L\in\mathbb{R}$. Note that in the following proof, despite using $L$, I have not assumed that $L$ is necessarily the limit of the sequence.
First, note the fact that $(x<y) \wedge (x\leq z) \implies z \leq y$, which we use in the proof. I will use an asterisk to make it clear when I have used this theorem.
Secondly, we have that $|s_n - s_m| = |s_n - s_m + L - L|$.
Now, for the first triangle inequality, we have
$$|(s_n - L) + (L-s_m)| \leq |s_n - L| + |s_m - L|,$$
which follows from the triangle inequality. We also have by hypothesis that $|s_n - s_m| < \epsilon$; thus* $|s_n - L| + |s_m - L| \leq \epsilon$.
Secondly, we have
$$ |s_n - L| \leq |(s_n - L) - (s_m - L)| + |s_m - L| \Leftrightarrow |s_n - L| - |s_m -L| \leq |s_n - s_m| < \epsilon$$
Which gives
$$|s_n - L| - |s_m -L| < \epsilon$$
We now have two inequalities,
$$ |s_n - L| - |s_m -L| < \epsilon \\ |s_n - L| + |s_m - L| \leq \epsilon$$
Which together imply that $$2|s_n - L| < 2\epsilon \implies |s_n - L| < \epsilon$$.
But it cannot possibly be the case that this is true for all $L\in\mathbb{R}$?
Have I made an error when deriving the fact that $(x<y) \wedge (x\leq z) \implies z \leq y$? Or have I made an error with the triangle inequality?
Edit: Obviously every convergent sequence is bounded, which proves the theorem.
The "fact" $(x<y) \wedge (x\leq z) \implies z \leq y$ which you have used is not true. For instance, $1<2$ and $1\leq 3$ but $3\leq 2$ is false.
You can also see directly that the way you used this fact in your argument doesn't make sense. You used it to deduce that since $s_n$ and $s_m$ are close to each other, $|s_n-L|+|s_m-L|$ must be small. That is clearly wrong, since $L$ could be far away from both $s_n$ and $s_m$ so $|s_n-L|$ and $|s_m-L|$ would be large.