Eigenspaces are subspaces

Identification. $Av=\lambda v\iff(A-\lambda I)v=0\iff v\in\ker(A-\lambda I)$.

Subspace. The kernel of a linear map is a subspace. Directly: $0\in E_\lambda$; if $v,w\in E_\lambda$, then $(A-\lambda I)(v+w)=0+0=0$; if $\alpha\in\mathbb{R}$ and $v\in E_\lambda$, then $(A-\lambda I)(\alpha v)=\alpha\cdot0=0$.

Eigenvalue criterion. By definition, $\lambda$ is an eigenvalue iff there is a nonzero $v$ with $Av=\lambda v$, i.e. iff $E_\lambda$ contains a nonzero vector.

Conclusion. $E_\lambda$ is a subspace, called the eigenspace of $A$ for $\lambda$; the eigenvectors for $\lambda$ are its nonzero elements. (The set of eigenvectors alone is not a subspace, since it excludes $0$.) If $\lambda$ is not an eigenvalue, $E_\lambda=\{0\}$.