Let:
$\mathbf{f}:D\subseteq \mathbb{R}^{n+1} \to \mathbb{R}^{n} \ \ \ D \ \text{open}$
$\ \ \ \ \ \ (t,\mathbf{y}) \mapsto \mathbf{f}(t,\mathbf{y})$
Where $t\in \mathbb{R}$ and $\mathbf{y} \in \mathbb{R}^{n}$.
$\mathbf{f}$ is locally Lipschitz in $D$ with respect to $\mathbf{y}$ and uniformly respect $t$ if and only if: $$\forall (t,\mathbf{y})\in D \ \exists B_r(t,\mathbf{y}) : \exists L\in\mathbb{R}_+ : \ \forall (t,\mathbf{z}) \in B_r(t,\mathbf{y}) \\ ||\mathbf{f}(t,\mathbf{y})-\mathbf{f}(t,\mathbf{y})||\leq L||\mathbf{y}-\mathbf{z}||$$ Where $L$ doesn't depend upon $t$.
The following proof is left as exercise to the reader
Let $\mathbf{f}(t,\mathbf{y})=(f_1(t,\mathbf{y}),...,f_n(t,\mathbf{y}))$ and $\mathbf{y}=(y_1,...,y_n)$ then: $$\mathbf{f}\text{ and }\frac{\partial f_j}{\partial y_s } \ \text{continous in } D \ \ \forall j,s\in\{1,...,n\} \implies \mathbf{f} \text{ is locally Lipschitz in }D\text{ with respect to }\mathbf{y}\text{ and uniformly respect }t $$
I'm trying to prove it. Here is my attempt:
Lemma
Since mean value theorem is not true for vectorial functions, my idea was to use this weaker inequality:
$\mathbf{g}:A\subseteq \mathbb{R}^n \to \mathbb{R}^n \ \ \ \mathbf{g}\in C^1(A)$
$\text{The segment } [\mathbf{y},\mathbf{z}]\subset A$
$\implies ||\mathbf{g}(\mathbf{y})-\mathbf{g}(\mathbf{z})||\leq \sqrt{n}\alpha||\mathbf{y}-\mathbf{z}||$
Where $$\alpha=\max \left\{\max_{[\mathbf{y},\mathbf{z}]} (||\nabla g_1(\mathbf{x})||),\max_{[\mathbf{y},\mathbf{z}]} (||\nabla g_2(\mathbf{x})||),...,\max_{[\mathbf{y},\mathbf{z}]} (||\nabla g_n(\mathbf{x})||)\right\}$$ The existences of the maximum of the $||\nabla g_j||$ is granted by Weierstrass Theorem, since $[\mathbf{y},\mathbf{z}]$ is compact.
Attempt of proof
I can define a family of functions $\mathbf{g}_t: \mathbf{y} \mapsto \mathbf{f}(t,\mathbf{y})$(I'll indicate $g_{t,j}$ its $j$-th component). One can easily verify the following:
$$ \nabla f_j(t_0,\mathbf{y}_0)=(D_t f_j(t_0,\mathbf{y}_0),\nabla g_{t_0,j}(\mathbf{y}_0)) \ \ \forall(t_0,\mathbf{y}_0)\in D $$
In fact:
$$D_{y_s} f_j(t_0,\mathbf{y}_0)=\lim_{h \to 0} \frac{f_j(t_0,\mathbf{y}_0+h\mathbf{e}_s)-f_j(t_0,\mathbf{y}_0)}{h}=\lim_{h \to 0} \frac{g_{t_0,j}(\mathbf{y}_0+h\mathbf{e}_s)-g_{t_0,j}(\mathbf{y}_0)}{h}=D_{y_s} g_{t_0,j}(\mathbf{y}_0)$$
This means that the following implication holds: $$D_{y_s} f_j(t,\mathbf{y})\leq N \ \ \ \forall(t,\mathbf{y})\in K\subset D \implies ||\nabla g_{t,j}(\mathbf{y})||\leq M \ \ \ \forall(t,\mathbf{y})\in K $$ In particular the LHS proposition is true for every $j$(and this obviously makes true also RHS for every $j$) by Weierstrass theorem if I choose $K$ to be a closed ball of generic center $(\overline{t},\mathbf{\overline{y}})$(this ball exists for every choice of $(\overline{t},\mathbf{\overline{y}})$, since $D$ is open).
After this premises, since $D_{y_s} \mathbf{g}_t:t \mapsto D_{y_s} \mathbf{f}(t,\mathbf{y}) $ and thanks to our hyphothesis it's easy to see that:
$\mathbf{g}_t:A_t\subseteq \mathbb{R}^n \to \mathbb{R}^n \ \ \ A_t=\{\mathbf{y}:(t,\mathbf{y})\in D \} $
$\mathbf{g}_t \in C^1(A_t)$
Moreover since a ball is connected $\forall (t,\mathbf{z})\in K \ \ $ the segment $[ (\overline{t},\overline{\mathbf{y}}),(t,\mathbf{z})]\subseteq K $ and this implies that $[\mathbf{\overline{\mathbf{y}}},\mathbf{z}]\subseteq A_t$(this can be easily proved showing that the pythagorean distance of $(\overline{t},\overline{\mathbf{y}})$ from $(t,q\overline{\mathbf{y}}+(1-q)\mathbf{z})$ with $q\in [0,1]$ is $\leq$ than the distance $d((\overline{t},\overline{\mathbf{y}}),(t,\mathbf{z}))$) , so by our lemma: $$\forall (t,\mathbf{z})\in K \ \ ||\mathbf{g}_t(\mathbf{\overline{y}})-\mathbf{g}_t(\mathbf{z})||\leq \sqrt{n}\alpha_t||\mathbf{\mathbf{\overline{y}}}-\mathbf{z}||$$
If we prove that $\alpha_t$ is bounded we are done because we substitute it with its upper bound, so that the inequality doesn't depend upon $t$.Notice that:
$$\alpha_t=\max \left\{\max_{[\mathbf{\overline{y}},\mathbf{z}]} (||\nabla g_{t,1}(\mathbf{y})||),\max_{[\mathbf{\overline{y}},\mathbf{z}]} (||\nabla g_{t,2}(\mathbf{y})||),...,\max_{[\mathbf{\overline{y}},\mathbf{z}]} (||\nabla g_{t,n}(\mathbf{y})||)\right\}$$
Since we know that for every $j \ \ $, $||\nabla g_{t,j}(\mathbf{y})||\leq M \ \ \forall (t,\mathbf{y})\in K$ ,we have just to prove that $\forall \mathbf{y} \in [\mathbf{\overline{y}},\mathbf{z}], (t,\mathbf{y})\in K$ , this can be done particularly easily showing that $ d((t,\mathbf{y}),(\overline{t},\mathbf{\overline{y}}))\leq d((t,\mathbf{z}),(\overline{t},\mathbf{\overline{y}}))$ (it's straight forward after you use the fact that $\mathbf{y}$ is convex combination of $\mathbf{\overline{y}}$ and $\mathbf{z}$). In conclusion: $$\alpha_t \leq M$$ And this implies that: $$\forall (t,\mathbf{z})\in K \ \ ||\mathbf{g}_t(\mathbf{\overline{y}})-\mathbf{g}_t(\mathbf{z})||\leq \sqrt{n}M||\mathbf{\mathbf{\overline{y}}}-\mathbf{z}||$$ By the definition of $\mathbf{g}_t$ and defining $L=M \sqrt{n}$: $$\forall (t,\mathbf{z})\in K \ \ ||\mathbf{f}(t,\mathbf{\overline{y}})-\mathbf{f}(t,\mathbf{z})||\leq L||\mathbf{\mathbf{\overline{y}}}-\mathbf{z}||$$ Since $(\overline{t},\overline{\mathbf{y}})$ was a generic point of $D$, this concludes the proof.
Is it correct?
It is correct, but a bit messy to read. In exchange for the first three formulas you write, you could simply state that, by definition, $$ \nabla f_j(t,\cdot) = \nabla g_{t,j}. $$ You also never define $N,M$ or $K$ which is bad form. You also should mention explicitly where you use the assumptions that are necessary for this to hold, such as the continuity of the partial derivatives, which I can't really find here. Here's how I would write up the proof (using your notation):
Fix $(t_0,\mathbf{y}) \in D$. Choose $r>0$ such that $B_r (t_0,\mathbf{y}) \subseteq D$, which is possible as $D$ is open. Defining $\mathbf{g}_t$ as you did, we have $$ \nabla f_j(t,\cdot) = \nabla g_{t,j}. $$ As $B_r(\mathbf{y})$ is compact, and the partial derivatives $\frac{\partial{f}_j}{\partial y_s}$ are continuous on $B_r(t_0,\mathbf y)$, it follows that there is a constant $\alpha$ such that $$ \sup_{(t,\mathbf x) \in B_r(t_0,\mathbf y)} \left|\frac{\partial{f}_j}{\partial y_s}(t,\mathbf x)\right| < \alpha, \quad j=1,\dots,n.$$ In particular, this means that for $(t,\mathbf x),(t,\mathbf z) \in B_r(t_0,\mathbf y)$, $$ \alpha_t=\max \left\{\max_{[\mathbf{x},\mathbf{z}]} (||\nabla g_{t,1}(\mathbf{y})||),\max_{[\mathbf{x},\mathbf{z}]} (||\nabla g_{t,2}(\mathbf{y})||),...,\max_{[\mathbf{x},\mathbf{z}]} (||\nabla g_{t,n}(\mathbf{y})||)\right\} < \alpha. $$ where we use that if $(t,\mathbf{x}), (t,\mathbf z) \in B_r(t_0,\mathbf y)$, any segment $\{t\} \times [\mathbf{x}, \mathbf{z}]$ lies in $B_r(t_0,\mathbf y)$ Applying the Lemma as you propose, we then have $$ \forall (t,z),(t,\mathbf{z})\in B_r(t_0,\mathbf{y}): \ \ ||\mathbf{g}_t(\mathbf{x})-\mathbf{g}_t(\mathbf{z})||\leq \sqrt{n}\alpha_t||\mathbf{x}-\mathbf{z}|| \leq \sqrt{n}\alpha||\mathbf{x}-\mathbf{z}||. $$ Choosing $L := \sqrt{n}\alpha$ then yields the claim.