Consider a $2 \times 2$ matrix $A$ with a single eigenvalue $\lambda$. For the eigenspace associated with $\lambda$ to be dimension two, $(\lambda I-A)$ must be the zero matrix (you need two independent parameters). So this is only possible if $A=\lambda I$.
Let $A \in \mathbb{R}^{n \times n}$ as you describe. In particular, we assume the complexified map $T(v)=Av$ for $v \in \mathbb{C}^n$ has $k$-fold distinct real eigenvalues. Of course, $T$ over $\mathbb{C}$ has $n$-complex eigenvalues as $\text{det}(T-tI)=0$ has $n$-zeros since $\mathbb{C}$ is algebraically closed. Continuing, we assume the remaining $(n-k)$ eigenvalues are also distinct and complex and arranged in $(n-k)/2$ conjugate pairs which I will denote $\alpha_j \pm i\beta_j$ for $j=1,2,\dots (n-k)/2$. Observe, over $\mathbb{C}^n$ this complexified map has $n$-distinct eigenvalues hence $T$ is diagonalizable. Let $\gamma = \{ v_1,\dots ,v_k, w_1, \bar{w}_1 \dots , w_{(n-k)/2}, \bar{w}_{(n-k)/2} \}$ be the complex eigenbasis which diagonalizes $T$. In particular, for $j=1,\dots , k$ we simply have:
$$ T(v_j) = \lambda_j v_j $$
then, as $\bar{A}=A$ we have $Aw=(\alpha+i\beta)w$ conjugated yields $A\bar{w}=(\alpha-i\beta)\bar{w}$. This shows the familiar result that, for real matrices $A$, complex eigenvectors come in conjugate pairs with conjugate eigenvalues. This is a beautiful and useful analog of the elementary pairing of complex roots to real polynomial equations. In particular, if $w=a+ib$ where $a,b \in \mathbb{R}^n$ and $\alpha,\beta \in \mathbb{R}$ such that:
\begin{align} T(w)=(\alpha+i\beta)w \ \ &\Rightarrow \ \ T(a+ib)=(\alpha+i\beta)(a+ib) \\
&\Rightarrow \ \ T(a)+iT(b)=\alpha a-\beta b + i(\beta a+\alpha b) \\
&\Rightarrow \ \ T(a)= \alpha a-\beta b \ \ \& \ \ T(b)= \beta a+\alpha b \\
\end{align}
Hence, we find an interesting real result. If we define $S$ to be the transformation on $\mathbb{R}^n$ with the same matrix $A$ as the complexified $T$ ( in other words, $S$ is the map with $[S]=A$ which we began this story) then
$$ S(a)= \alpha a-\beta b \ \ \& \ \ S(b)= \beta a+\alpha b $$
Let $W = \text{span}\{ a,b \}$ then $S|_W=S'$ has the following block structure with respect to basis $\omega = \{a,b \}$: (your text might use $[S']_{\omega }^{\omega}$)
$$ [S']_{\omega \omega} = \left[[S(a)]_{\omega} \ | \ [S(b)]_{\omega}\right] = [[\alpha a-\beta b]_{\omega} \ | \ [\beta a+\alpha b]_{\omega}] = \left[ \begin{array}{cc}\alpha & \beta \\ -\beta & \alpha \end{array} \right]$$
Therefore, to find the block-decomposition requested by the OP we simply use a basis for $\mathbb{R}^n$ assembled from $k$-eigenvectors $v_1, \dots v_k$ followed by the real and imaginary parts of the complex eigenvectors where we choose to make use of the first copy rather than the conjugate in each conjugate pair. In particular,
$$ \gamma = \{ v_1,\dots , v_k, a_1,b_1, \dots , a_m,b_m \} $$
where $m = (n-k)/2$ and $A(a_j+ib_j)=(\alpha_j+i\beta_j)(a_j+ib_j)$ for each $j=1, \dots , m$. It follows from the discussion given thus far:
$$ [\gamma]^{-1}A[\gamma] = \left[
\begin{array}{cccccccc}
\lambda_1 & & & & & & & \\
& \ddots & & & & & & \\
& & \lambda_k & & & & & \\
& & & \alpha_1 & \beta_1 & & & \\
& & & -\beta_1 & \alpha_1 & & & \\
& & & & & \ddots & & \\
& & & & & & \alpha_m & \beta_m \\
& & & & & & -\beta_m & \alpha_m \end{array}\right]$$
This is known as the real Jordan form of $A$. In the event that the complexified $A$ is not diagonalized the real Jordan form picks up more terms in the super diagonals from the $1$'s in the superdiagonal of the complex Jordan form of $A$. Philosophically, this discussion is at odds with those mathematicians who refuse to think across fields. There is a pun here I intend. One text which has this is Ralph Abraham's classic text Linear and Multilinear Algebra. Most of my other linear algebra texts have little on this topic as it is just cleaner to stick with either $\mathbb{R}$ or $\mathbb{C}$ etc. But, this concept of complexification is exceedingly important to properly understand applications of complex math to real problems and I think we would do well to spend more time on the exposition of this structure.
Best Answer
The matrix is obviously nilpotent, and in fact $\;A^3=0\,,\,\,A^2\neq0\;$ , and from here its JCF is
$$J_A=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix}$$
Or easier, as you began to argue: a nilpotent matrix is diagonalizable iff it is the zero matrix, so our matrix is non-diagonalizable...so now choose the size of the biggest Jordan block according to the rank of $\;A\;$ ...