Is there a proof or example to show that a noiseless subsystem is not necessarily closed under addition?

I do not understand how if some quantum states $rho, sigma in mathbb{C}^n otimes mathbb{C}^m$ then how can their addition not be in $mathbb{C}^n otimes mathbb{C}^m$?

The author is not claiming that this is false. This is certainly true, you are right about that!

In a [text][1] (section 3.6 page 92) about noiseless subsystems by D. Lidar, it is mentioned:'A subsystem is a tensor factor in a tensor product, and this does not have to be a subspace (e.g., in general it is not closed under addition).'

The author is saying that there are non-subspace structures (i.e. tensor factors) which have the noiseless property. The point is kind of abstract but it comes from the notion that $mathbb{C}^n$ isn't naturally a subspace of $mathbb{C}^notimes mathbb{C}^m$. However, this doesn't mean you can't think of a way to think of $mathbb{C}^n$ as a subspace, it just means you need to fix something in $mathbb{C}^m$ to do so.

More to the point, if you did want to think about $mathbb{C}^n$ as a subspace of $mathbb{C}^notimes mathbb{C}^m$ you would need to fix a basis of $mathbb{C}^m$ and identify how the basis of $mathbb{C}^n$ gets mapped to the basis of $mathbb{C}^notimes mathbb{C}^m$. It's because of this "identication requirement" these noiseless subsystems are not technically subspaces.

Ultimately, the difference between noiseless subspace and noiseless subsystem is subtle, but I hope to illustrate the difference with an example. Encoding a single qubit into a noiseless subspace of a 2-qubit space can be given by the mapping $$alpha|0rangle+beta|1rangle mapsto alpha|0rangle_1|0rangle_2+beta|1rangle_1|1rangle_2$$ this encoding can be seen as a linear (or subspace) endcoding as we are mapping a basis of $mathbb{C}^2$ to a basis of $mathbb{C}^2otimes mathbb{C}^2cong mathbb{C}^4$. Now, consider an encoding of a qubit of the form $$alpha|0rangle+beta|1rangle mapsto (alpha|0rangle_1+beta|1rangle_1)otimes |psirangle_2 $$ where $|psirangle_2$ is an arbitrary qubit in the second space of $mathbb{C}^2otimes mathbb{C}^2$. This encoding is into the first subsystem and is not a subspace encoding because we have not fixed the state of the second system in the tensor product (this makes the encoding non-linear and that's why its not the same as a subspace encoding). That being said, if we did fix $|psirangle_2$ to be a specific element of $mathbb{C}^2$ then we would get a subspace encoding like the first example, just in a different basis.