Suppose $\sum_{n=1}^\infty \frac{1}{n} = S$ where $S$ is finite. Then
$S =\sum_{n=1}^\infty \frac{1}{n}= \sum_{n=1}^\infty \frac{1}{2n-1} + \frac{1}{2n} > \sum_{n=1}^\infty \frac{1}{2n} + \frac{1}{2n} = S$
which is a contradiction. Is this valid?
Suppose $\sum_{n=1}^\infty \frac{1}{n} = S$ where $S$ is finite. Then
$S =\sum_{n=1}^\infty \frac{1}{n}= \sum_{n=1}^\infty \frac{1}{2n-1} + \frac{1}{2n} > \sum_{n=1}^\infty \frac{1}{2n} + \frac{1}{2n} = S$
which is a contradiction. Is this valid?
One way to express your argument more formally is by considering the partial sums
$S_{2m}=\sum_{n=1}^{2m}\frac1n=\sum_{n=1}^m\left(\frac1{2n-1}+\frac1{2n}\right)\;.$
You can bound these partial sums as you did to obtain
$ S_{2m}\gt\sum_{n=1}^m\left(\frac1{2n}+\frac1{2n}\right)=\sum_{n=1}^m\frac1n=S_m\;.$
If the series were to converge, the sequences $S_{2m}$ and $S_m$ would have to converge to a common limit, so their difference would have to converge to $0$ for $m\to\infty$, whereas in fact the gap between them increases with $m$.