I am reading this text:
Is my understanding of this correct? Because any two vectors added together are still a linear combination of vectors in S, then it is closed under addition. Any two vectors added together are still in span(s), so it is closed under addition, and therefore span(s) is still a subspace because one principle of being a subspace is that it is closed under addition.
Yes, it is correct. You could add to that that iv $v\in\operatorname{span}S$ and $\alpha$ is a scalar, then $\alpha v\in\operatorname{span}S$ too. So, $\operatorname{span}S$ is closed both under addition and multiplication by scalars.