Show that $\lim_{n}\sum_{k=n}^{2n}{1\over k} = \ln2$ using elementary methods.

Your argument is ok if you take $(1) $ as the starting point. I would say that your writing is too convoluted, and you are misusing the $\exists $ symbol.

You could simply say \begin{align} \sum_{k=n+1}^{2n}\frac1k-\log2=\left (\sum_{k=1}^{2n}\frac1k-\log2n\right)-\left (\sum_{k=1}^{n}\frac1k-\log n\right)\xrightarrow [n\to\infty]{}L-L=0. \end{align} You need to distinguish between how you get the idea, and how you write the proof.