If $\{x_n\}$ is an increasing sequence and $\lim_{n\to\infty}x_n=L$, then $L$ is an upper bound of $\{x_n\}$

This is a very clear proof, easy and natural to follow. The only 'improvement' that I could see would be that it's possible to compactify it a lot, as in squeeze the information into a paragraph or less, with the upside being space advantage and the downside being reduced readability. All in all though, this proof is clear and concise which is pretty much what everybody should aim for.

Your proof is fine.

Slightly different:

Given: $x_n$ is increasing, convergent to $L$.

Need to show that $x_n \le L$,

Assume there is a $n_0 \in \mathbb{N}$ s.t.

$L < x_{n_0}$.

For $n \ge n_0$ :

$x_{n_0} \le x_n$ since $x_n$ is increasing.

But then

$L <x_{n_0} \le \lim_{ n \rightarrow \infty}x_n = L$,

a contradiction.