Showing following Riemann Zeta series converges for $\sigma>1$ is considered easy in many sources. Here $s=\sigma+it$.
$$\sum \frac{1}{n^s}$$
A previous post presents the following argument [link].
$$\left \vert \displaystyle \sum \dfrac1{n^s} \right \vert \leq \displaystyle \sum \left \vert \dfrac1{n^s} \right \vert = \displaystyle \sum \left \vert \dfrac1{n^{\sigma + it}} \right \vert = \displaystyle \sum \dfrac1{n^{\sigma}} \left \vert \dfrac1{n^{it}} \right \vert = \displaystyle \sum\dfrac1{n^{\sigma}}$$
That is,
$$\left \vert \sum \dfrac1{n^s} \right \vert \leq \sum\dfrac1{n^{\sigma}}$$
The argument is that since $\sum 1/n^{\sigma}$ converges for $\sigma>1$ (by comparison with integrals for example) so does $\sum 1/n^{s}$.
Question: Surely this argument only tells that $|\sum 1/n^{s}|$ is bounded, not necessarily convergent, for $\sigma>1$.
You are missing the theorem that states that if a series converges absolutely, then it converges. If $\sum|a_n|$ converges then $\sum a_n$ converges.
And to prove convergence of $\sum|a_n|$ it's enough to prove it's bounded since it's a series of positive reals.