Why an eigenspace is a linear subspace, if the zero vector is not an eigenvector?

You have two solutions to this.

Either you call a non-zero $v \in V$ (your vector space) an eigenvector of $A : V \to V$ if and only if there exists $\lambda$ such that $Av = \lambda v$, in which case you say $\lambda$ is an eigenvalue of $A$ associated to $v$.

Or

You call $\lambda$ an eigenvalue of $A$ if $\dim \ker(A - \lambda \mathrm{id}) > 0$, and you define eigenvectors associated to the eigenvalue $\lambda$ as non-zero elements of $\ker(A-\lambda \mathrm{id}_V)$.

In both cases you have to exclude $0$ as an eigenvector in some way. Also in both cases, the eigenspace is defined as $\ker(A- \lambda \mathrm{id}_V)$, so of course $0$ is naturally included in it.

Hope that helps,

The eigenspace associated with an eigenvalue consists of all the eigenvectors (which by definition are not the zero vector) associated with that eigenvalue along with the zero vector.

If we allowed the zero vector to be an eigenvector, then every scalar would be an eigenvalue, which would not be desirable.