Let $f:\mathbb{R}^{n}\times \mathbb{R}^{n}\to \mathbb{R}$ be of class $C^{k+2}$, $k \ge 0$ and suppose that, for each $(x,y) \in \mathbb{R}^{n}\times \mathbb{R}^{n}$, the matrix: $$\bigg{[}\frac{\partial^{2}f}{\partial y_{i}\partial y_{j}}(x,y)\bigg{]}_{1\le i,j\le n}$$ is positive. In other words, for each fixed $x \in \mathbb{R}^{n}$, the mapping $y \mapsto f_{x}(y) = f(x,y)$ is strictly convex.
For each fixed $x \in \mathbb{R}^{n}$ and $p \in \mathbb{R}^{n}$, define $\varphi_{x,p}: \mathbb{R}^{n}\to \mathbb{R}$ by: $$\varphi_{x,p}(y) = \langle y, p \rangle - f(x,y).$$
It follows from the hypothesis on $f$ that $\varphi_{x,p}$ has at most one critical point and, any such critical point (if it exists) is a global maximum.
I want to prove the following:
Proposition: The set $V =\{(x,p)\in \mathbb{R}^{n}\times \mathbb{R}^{n}: \mbox{$\varphi_{x,p}$ has a global maximum}\}$ is nonempty and open.
Attempted Proof: Note that $(x,p) \in V \iff \exists y_{0} \in \mathbb{R}^{n}$ such that $\nabla_{y}\varphi_{x,p}(y_{0}) = 0 \iff \nabla_{y}f(x,y_{0}) = p$. Hence: $$V = \{(x,p) \in \mathbb{R}^{n}\times \mathbb{R}^{n}: \mbox{$ y_{0} \in \mathbb{R}^{n}$ such that $\nabla_{y}f(x,y_{0}) = p$}\}$$
For each $x \in \mathbb{R}^{n}$ fixed, $\nabla_{y}f(x,\cdot)$ is injective and, hence, invertible. Let $p$ be an element of its range. In this case, there exists a unique $y_{0} \in \mathbb{R}^{n}$ with $\nabla_{y}f(x,y_{0}) = p$, so that $V \neq \emptyset$.
My question is: how to prove $V$ is open? I sketched the following proof. Let $V_{x} = \{p \in \mathbb{R}^{n}: \mbox{$p=\nabla_{y}f(x,y_{0})$ for some $y_{0}$}\}$. This is open because it is the pre-image of $\nabla_{y}f(x,\cdot)$. Now $V = \bigcup_{x\in \mathbb{R}^{n}}\{x\}\times V_{x}$. But is this set open?
Note: the notation $\nabla_{y}f(x,y)$ means $(\frac{\partial f}{\partial y_{1}}(x,y),...,\frac{\partial f}{\partial y_{n}}(x,y))$.
Your idea is on the right track, but the set $V$ will be open if you can show that for every point in $V$, there exists an open neighborhood around that point that lies entirely within $V$. In other words, you need to show that if $(x_0, p_0) \in V$, then there exists an open neighborhood $U$ of $x_0$ and an open neighborhood $W$ of $p_0$ such that for every $(x, p) \in U \times W$, we have $(x, p) \in V$.
Let's suppose that $(x_0, p_0) \in V$, so by definition of $V$ there exists $y_0 \in \mathbb{R}^n$ such that $\nabla_y f\left(x_0, y_0\right) = p_0$. Then consider the following two steps:
Find a neighborhood of $x_0$: Because the second derivative of $f$ with respect to $y$ is continuous (since $f$ is $C^{k+2}$), there exists a neighborhood $U$ of $x_0$ such that $\left\|\nabla_y^2 f(x, y_0) - \nabla_y^2 f(x_0, y_0)\right\| < \varepsilon$ for all $x \in U$, where $\varepsilon > 0$ is a small number (which will be chosen later).
Find a neighborhood of $p_0$: The function $\nabla_y f$ is smooth and therefore locally Lipschitz continuous. This means that there exists a neighborhood $W$ of $p_0$ and a constant $L > 0$ such that for all $p \in W$, we have $\|\nabla_y f(x, y) - p\| \leq L \|y - y_0\|$ for all $x \in U$, where $U$ is the neighborhood of $x_0$ from step 1.
Now, take $\varepsilon$ in step 1 to be small enough that $L \varepsilon < 1$. Then for every $(x, p) \in U \times W$, the map $y \mapsto \langle y, p \rangle - f(x, y)$ is strictly convex and its gradient is close to the gradient at $(x_0, y_0)$, so it has a unique critical point, which is a global maximum. Hence, $(x, p) \in V$.
This shows that $V$ is open, as needed.