Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $A: D(A) \subset H \to H$ be a maximal monotone (unbounded linear) operator. We define by induction the subspaces $$ D(A^{k+1}) := \{ v \in D(A^k) :Av \in D(A^k)\} \quad \forall k \in \mathbb N^*. $$ Then $D(A^k)$ is a Hilbert space for the inner product $$ \langle u, v \rangle_{D(A^k)} := \sum_{j=0}^k \langle A^j u, A^j v \rangle. $$
Then we have from Brezis' Functional Analysis two theorems, i.e.,
Theorem 7.4 (Hille-Yosida) Let $A: D(A) \subset H \to H$ be a maximal monotone (unbounded linear) operator. Then, given any $u_0 \in D(A)$ there exists a unique function $$ u \in C^1([0,+\infty) ; H) \cap C([0,+\infty) ; D(A)) $$ satisfying $$ (*) \quad \begin{cases} \frac{d u}{d t}+A u&=0 \quad \text{on} \quad [0,+\infty), \\ u(0)&=u_0. \end{cases} $$ Moreover, $$ |u(t)| \leq\left|u_0\right| \quad \text { and } \quad\left|\frac{d u}{d t}(t)\right|=|A u(t)| \leq\left|A u_0\right| \quad \forall t \geq 0. $$
and
Theorem 7.5 Assume $u_0 \in D (A^k)$ for some integer $k \geq 2$. Then the solution $u$ of problem $(*)$ obtained in Theorem 7.4 satisfies $$ u \in C^{k-j} ([0,+\infty) ; D (A^j)) \quad \forall j=0,1, \ldots, k. $$
The author proves Theorem 7.5 by first considering the case $k=2$ and then by induction for $k \ge 3$. From a comment by @Kroki, I would like to treat the cases $k=2$ and $k \ge 3$ simultaneously.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
We argue by induction on $k \ge 2$: assume that the result holds up to order $(k-1)$.
Consider the Hilbert space $H_1 = D(A^{k-1})$ equipped with the inner product $\langle \cdot, \cdot \rangle_{D(A^{k-1})}$. Then the operator $A_1 :D(A_1) \subset H_1 \to H_1$ defined by $$ \begin{cases} D(A_1) &= D(A^k), \\ A_1 u &= Au, \end{cases} $$ is maximal monotone.
We have $u_0 \in D(A_1)$. We apply Theorem 7.4 to the operator $A_1$ in the space $H_1$, we see that there is a unique function $$ u \in C^1([0,+\infty) ; H_1) \cap C([0,+\infty) ; D(A_1)) $$ such that $$ \begin{cases} \frac{d u}{d t}+A u &= 0 \quad \text{on} \quad [0,+\infty), \\ u(0) &= u_0. \end{cases} \tag{1} $$
Notice that $H_1 \subset H$ and $D(A_1) \subset D(A)$. Then $$ u \in C^1([0,+\infty) ; H) \cap C([0,+\infty) ; D(A)) $$
It follows from (1) that $u$ satisfies $(*)$. By uniqueness, $u$ is the solution of $(*)$.
Since $A \in \mathcal L(H_1, H)$ and $u \in C ([0, +\infty); H_1)$, we get $Au \in C^1 ([0, +\infty); H)$ and $$ \frac{d}{dt} (A u) = A \left (\frac{d u}{dt} \right ) \quad \text{on} \quad [0, +\infty). $$
By $(*)$, $\frac{d u}{d t} \in C^1 ([0, +\infty); H)$ and thus $u \in C^2 ([0, +\infty); H)$. Also, $$ \frac{d}{dt} \left (\frac{d u}{d t} \right )+A \left (\frac{d u}{dt} \right )=0 \quad \text{on} \quad [0,+\infty). $$
Let $$ v :=\frac{d u}{d t}. $$
We have just shown above that $$ v \in C^1 ([0, +\infty); H) \cap C([0, +\infty); H_1), $$ and $$ \begin{cases} \frac{d v}{d t}+A v &= 0 \quad \text{on} \quad [0,+\infty), \\ v(0) &= -A u_0. \end{cases} $$
Then $v$ is the solution of $(*)$ with the initial data $v(0) = -A u_0$. By inductive hypothesis, $$ v \in C^{k-1-j} ([0,+\infty) ; D (A^j)) \quad \forall j=0,1, \ldots, k-1. \tag{2} $$
This implies $$ u \in C^{k-j} ([0,+\infty) ; D (A^j)) \quad \forall j=0,1, \ldots, k-1. $$
It remains to check that $$ u \in C ([0,+\infty) ; D(A^k)). \tag{3} $$
Applying (2) with $\ell=k-1$, we get $$ \frac{du}{dt} \in C ([0,+\infty) ; D(A^{k-1})). $$
Then by $(*)$, $$ Au \in C ([0,+\infty) ; D(A^{k-1})). $$
Notice that $$ |v|_{D(A^k)}^2 = |v|^2 + |Av|_{D(A^{k-1})}^2 \quad \forall v \in D(A^k). $$
Then (3) follows. This completes the proof.