I came across the following easy fact (Lemma 3.17) in Meiss' Differential Dynamical Systems. Here is my paraphrased version:
Lemma. Let $E \subseteq \mathbb{R}^d$, let $I \subset \mathbb{R}$ be a compact interval, let $f \in C^k(E,\mathbb{R}^d)$ for some $k \geq 0$, and suppose that $x: I \to \mathbb{R}^d$ solves the IVP $$\dot{x}(t) = f(x(t)), \; x(0) = x_0.$$ Then $x \in C^{k+1}(I,\mathbb{R}^d)$.
I am trying to use this lemma to prove the following analogous claim for nonautonomous IVPs:
Claim. Let $E$ and $I$ be as in the lemma, let $f \in C^k(E \times I, \mathbb{R}^d)$, and suppose that $x: I \to \mathbb{R}^d$ solves the IVP \begin{equation} \dot{x}(t) = f(x(t),t), \quad x(0) = x_0. \end{equation}
Then $x \in C^{k+1}(I,\mathbb{R}^d).$
Here is my proof attempt.
Proof. Firstly, we can write $x(t) = (x_1(t),\ldots,x_d(t))$ and $f(x,t) = (f_1(x,t),\ldots,f_d(x,t))$, where $x_i: I \to \mathbb{R}$ and $f_i: \mathbb{R}^d \times I \to \mathbb{R}$ for each $i=1,\ldots,d$. Now define $\tilde{x}: I \to \mathbb{R}^{d+1}$ by \begin{equation} \tilde{x}(t) := (x_1(t),\ldots,x_d(t),t) \qquad \forall t \in I. \end{equation}
Then for each $i = 1,\ldots,d$, define $\tilde{f}_i: \mathbb{R}^{d+1} \to \mathbb{R}$ by \begin{equation} \tilde{f}_i(s_1,\ldots,s_d,s_{d+1}) := f_i((s_1,\ldots,s_d),s_{d+1}) \end{equation}
for all $s = (s_1,\ldots,s_{d+1}) \in \mathbb{R}^{d+1}$. Now note that \begin{align*} \tilde{f}_i(\tilde{x}(t)) &= \tilde{f}_i(x_1(t),\ldots,x_d(t),t) \\[3pt] &= f_i((x_1(t),\ldots,x_d(t)),t) \\[3pt] &= f_i(x(t),t). \end{align*}
Next, define $\tilde{f}: \mathbb{R}^{d+1} \to \mathbb{R}^{d+1}$ by \begin{equation} \tilde{f}(s) := (\tilde{f}_1(s),\ldots,\tilde{f}_d(s),1) \qquad \forall s \in \mathbb{R}^{d+1}. \end{equation}
Clearly, $\tilde{f} \in C^k(\tilde{E},\mathbb{R})$, where $\tilde{E} := E \times I$. Let $x_0 := (x_0^{(1)},\ldots,x_0^{(d)})$ and define $\tilde{x}_0 := (x_0^{(1)},\ldots,x_0^{(d)},0)$. Now observe that $\tilde{x}$ solves the autonomous IVP \begin{equation} \dot{\tilde{x}}(t) = \tilde{f}(\tilde{x}(t)), \quad \tilde{x}(0) = \tilde{x}_0, \end{equation}
since \begin{align*} \dot{\tilde{x}}(t) &= (\dot{x}_1(t),\ldots,\dot{x}_d(t),1) \\[3pt] &= (f_1(x(t),t),\ldots,f_d(x(t),t),1) \\[3pt] &= (\tilde{f}_1(\tilde{x}(t)),\ldots,\tilde{f}_d(\tilde{x}(t)),1) \\[3pt] &= \tilde{f}(\tilde{x}(t)). \end{align*}
It then follows by the lemma that $\tilde{x} \in C^{k+1}(\tilde{E},\mathbb{R})$, which then implies that $x \in C^{k+1}(E,\mathbb{R}). \qquad \square$
My main question is: Is this proof correct? I just want to make sure my reasoning is sound.
Update 7/16/23: Thanks to @Filippo for pointing out a much simpler proof:
Proof: Fix $k \geq 0$ and let $f \in C^k(E \times I,\mathbb{R}^d)$. We prove that $x \in C^j(I,\mathbb{R}^d)$ for all $j \in \{0,1,\ldots,k+1\}$ by induction on $j$.
Base case $(j = 0)$: By assumption $x$ is differentiable and hence and hence continuous, and so $x \in C^0(I,\mathbb{R}^d)$.
Inductive step: Fix an arbitrary $j \in \{0,1,\ldots,k\}$ and assume $x \in C^j(I,\mathbb{R}^d)$. Then the map $t \mapsto f(x(t))$ is in $C^j(I,\mathbb{R}^d)$ being that it's the composition of the maps $f \in C^j(E \times I,\mathbb{R}^d)$ and $t \mapsto (x(t),t) \in C^j(I,\mathbb{R}^d \times I)$. Hence, $\dot{x} \in C^j(I,\mathbb{R}^d)$ and so $x \in C^{j+1}(I,\mathbb{R}^d)$. This completes the proof.