Given $A\in M_{n,n}(\mathbb{R})$ and $\lambda$ an eigenvalue, a generalized eigenvector of rank $i$ is defined as $v \in ker(A-\lambda E)^i\setminus ker(A-\lambda E)^{i-1}$.
Why does such vector exist? Why is it that $\{ 0 \} \subset \ldots \subset ker(A-\lambda E)^{i-1}\subset ker(A-\lambda E)^i\subset \ldots \subset \mathbb{R}^n $?
Your second question is easier, so let's start with that. Note that, by definition, $$ \ker(A - \lambda E)^k = \{x \in \Bbb R^n : (A - \lambda E)^k x = 0\}. $$ We want to see that $\ker(A - \lambda E)^{k-1} \subset \ker(A - \lambda E)^{k}$. In other words, if $(A - \lambda E)^{k-1} x = 0$, then $(A - \lambda E)^k x = 0$.
So, suppose that $x \in (A - \lambda E)^{k-1}$, which is to say that $(A - \lambda E)^{k-1} x = 0$. It follows that $$ (A - \lambda E)^k x = (A - \lambda E)[(A - \lambda E)^{k-1} x] = (A - \lambda E)\ 0 = 0. $$
As for why such vectors exist: the key is to note that if $A$ is diagonalizable, then there exists a basis of eigenvectors of $A$. In other words, the subspaces $\ker (A - \lambda E)$ (that is, the eigenspaces of $A$) have dimensions that add up to $n$.
If $A$ is not diagonalizable, then this is no longer the case. However, the Cayley-Hamilton theorem tells us that $$ (A - \lambda_1 E)^{m_1} \cdots (A - \lambda_p E)^{m_p} = 0, $$ which allows us to deduce that the generalized eigenspaces $\ker(A - \lambda E)^m$ have dimensions that add up to $n$. In other words: there must be an eigenvalue $\lambda_j$ for which $\dim \ker (A - \lambda_j E)^{m_j}$ is bigger than $\dim \ker (A - \lambda_j E)$.