[proofplan]
We prove lower semicontinuity by testing an arbitrary sequence of perturbations $u_k \to 0$ in the relative domain. If the limiting inferior of the values is finite, we choose nearly minimizing points $x_k$ for the perturbed problems; the uniform compactness hypothesis forces these points into one compact set, so a convergent subsequence exists. Closedness of $\Phi$ then passes the limiting inequality to the limit problem at $u = 0$. The feasible point map gives the opposite inequality for the upper limit, so the marginal values converge to $p(0)$.
[/proofplan]
[step:Reduce lower semicontinuity to a finite liminf subsequence]
Let $(u_k)_{k=1}^{\infty}$ be a sequence in $U \cap \operatorname{dom} p$ such that $u_k \to 0$ in $\mathbb{R}^m$. Define
\begin{align*}
L := \liminf_{k \to \infty} p(u_k) \in [-\infty,+\infty].
\end{align*}
Since $u_k \in U \cap \operatorname{dom} p$ for every $k$, the local lower bound gives $p(u_k) \geq \beta$ for every $k$. Hence $L \geq \beta$, so $L \neq -\infty$.
If $L = +\infty$, then $p(0) \leq L$ because $p(0) \in \mathbb{R}$. It remains to consider the case $L \in \mathbb{R}$. By the definition of $\liminf$, there exists a strictly increasing map
\begin{align*}
j: \mathbb{N} \to \mathbb{N}
\end{align*}
such that
\begin{align*}
p(u_{j(k)}) \to L.
\end{align*}
Replacing the original sequence by the subsequence $(u_{j(k)})_{k=1}^{\infty}$, it is enough to prove
\begin{align*}
p(0) \leq \lim_{k \to \infty} p(u_k).
\end{align*}
[guided]
We begin with an arbitrary sequence of admissible perturbations. Let $(u_k)_{k=1}^{\infty}$ be a sequence in $U \cap \operatorname{dom} p$ with $u_k \to 0$. To prove relative lower semicontinuity, we must show
\begin{align*}
p(0) \leq \liminf_{k \to \infty} p(u_k).
\end{align*}
Define the extended real number
\begin{align*}
L := \liminf_{k \to \infty} p(u_k).
\end{align*}
The assumption $u_k \in U \cap \operatorname{dom} p$ puts every $u_k$ inside the region where the local lower bound applies, so
\begin{align*}
p(u_k) \geq \beta
\end{align*}
for every $k \in \mathbb{N}$. Therefore
\begin{align*}
L \geq \beta.
\end{align*}
This is the point of the lower bound hypothesis: it prevents the marginal values from tending to $-\infty$ along admissible perturbations near $0$.
If $L = +\infty$, then the desired inequality follows immediately because $p(0)$ is a finite real number. Thus the only substantive case is $L \in \mathbb{R}$. In that case, the definition of the lower limit gives a subsequence along which the values converge to the lower limit. More precisely, there is a strictly increasing map
\begin{align*}
j: \mathbb{N} \to \mathbb{N}
\end{align*}
such that
\begin{align*}
p(u_{j(k)}) \to L.
\end{align*}
The perturbations still converge to $0$ along this subsequence, because every subsequence of a convergent sequence has the same limit. Hence we may replace $(u_k)$ by $(u_{j(k)})$ and prove
\begin{align*}
p(0) \leq \lim_{k \to \infty} p(u_k).
\end{align*}
This proves the original lower semicontinuity inequality because the chosen subsequence realizes the lower limit.
[/guided]
[/step]
[step:Choose near-minimizers for the finite-value subsequence]
Assume now that $(u_k)_{k=1}^{\infty}$ is a sequence in $U \cap \operatorname{dom} p$ such that $u_k \to 0$ and
\begin{align*}
p(u_k) \to L \in \mathbb{R}.
\end{align*}
Define
\begin{align*}
\varepsilon_k := \frac{1}{k}
\end{align*}
for $k \in \mathbb{N}$. Then $\varepsilon_k > 0$ and $\varepsilon_k \downarrow 0$.
Because $u_k \in \operatorname{dom} p$, we have $p(u_k) < +\infty$. The lower bound gives $p(u_k) \geq \beta$, so $p(u_k) \in \mathbb{R}$. By the definition of the infimum
\begin{align*}
p(u_k) = \inf_{x \in \mathbb{R}^n} \Phi(x,u_k),
\end{align*}
there exists $x_k \in \mathbb{R}^n$ such that
\begin{align*}
\Phi(x_k,u_k) \leq p(u_k) + \varepsilon_k.
\end{align*}
Thus $(x_k)_{k=1}^{\infty}$ is a sequence of $\varepsilon_k$-near-minimizers for the perturbed marginal problems.
[/step]
[step:Extract a compactly convergent subsequence of near-minimizers]
The sequence $(u_k)_{k=1}^{\infty}$ lies in $U \cap \operatorname{dom} p$, satisfies $u_k \to 0$, and the sequence $(\varepsilon_k)_{k=1}^{\infty}$ satisfies $\varepsilon_k \downarrow 0$. The uniform compactness hypothesis therefore gives $k_0 \in \mathbb{N}$ such that
\begin{align*}
x_k \in K
\end{align*}
for every $k \geq k_0$.
Since $K \subset \mathbb{R}^n$ is compact, there exist a strictly increasing map
\begin{align*}
r: \mathbb{N} \to \mathbb{N}
\end{align*}
with $r(k) \geq k_0$ for every $k$, and a point $x^* \in K$, such that
\begin{align*}
x_{r(k)} \to x^*
\end{align*}
in $\mathbb{R}^n$. Since also $u_{r(k)} \to 0$, we have
\begin{align*}
(x_{r(k)},u_{r(k)}) \to (x^*,0)
\end{align*}
in $\mathbb{R}^n \times \mathbb{R}^m$.
[/step]
[step:Use closedness of $\Phi$ to pass the near-minimal inequality to the limit]
Because $\Phi$ is closed, it is lower semicontinuous on $\mathbb{R}^n \times \mathbb{R}^m$. Applying this lower semicontinuity to the convergent sequence $(x_{r(k)},u_{r(k)}) \to (x^*,0)$ gives
\begin{align*}
\Phi(x^*,0) \leq \liminf_{k \to \infty} \Phi(x_{r(k)},u_{r(k)}).
\end{align*}
For every $k \in \mathbb{N}$, the near-minimality inequality gives
\begin{align*}
\Phi(x_{r(k)},u_{r(k)}) \leq p(u_{r(k)}) + \varepsilon_{r(k)}.
\end{align*}
Taking lower limits and using $p(u_k) \to L$ and $\varepsilon_k \to 0$, we obtain
\begin{align*}
\liminf_{k \to \infty} \Phi(x_{r(k)},u_{r(k)}) \leq \lim_{k \to \infty} \bigl(p(u_{r(k)}) + \varepsilon_{r(k)}\bigr) = L.
\end{align*}
Combining the two inequalities yields
\begin{align*}
\Phi(x^*,0) \leq L.
\end{align*}
Since
\begin{align*}
p(0) = \inf_{x \in \mathbb{R}^n} \Phi(x,0),
\end{align*}
we also have
\begin{align*}
p(0) \leq \Phi(x^*,0).
\end{align*}
Therefore
\begin{align*}
p(0) \leq L.
\end{align*}
This proves that $p$ is lower semicontinuous at $0$ relative to $U \cap \operatorname{dom} p$.
[/step]
[step:Use the feasible point map to obtain the matching upper bound]
Assume now that there exist a neighbourhood $V \subset U$ of $0$ and a map
\begin{align*}
x: V \cap \operatorname{dom} p \to \mathbb{R}^n
\end{align*}
such that
\begin{align*}
\Phi(x(u),u) \to \Phi(x(0),0) = p(0)
\end{align*}
as $u \to 0$ through $V \cap \operatorname{dom} p$.
Let $(v_k)_{k=1}^{\infty}$ be any sequence in $V \cap \operatorname{dom} p$ such that $v_k \to 0$. Since $V \subset U$, the lower semicontinuity already proved gives
\begin{align*}
p(0) \leq \liminf_{k \to \infty} p(v_k).
\end{align*}
For each $k \in \mathbb{N}$, the definition of $p(v_k)$ as an infimum over $x \in \mathbb{R}^n$ gives
\begin{align*}
p(v_k) \leq \Phi(x(v_k),v_k).
\end{align*}
Taking upper limits and using the assumed convergence of the feasible values gives
\begin{align*}
\limsup_{k \to \infty} p(v_k) \leq \lim_{k \to \infty} \Phi(x(v_k),v_k) = p(0).
\end{align*}
Hence
\begin{align*}
p(0) \leq \liminf_{k \to \infty} p(v_k) \leq \limsup_{k \to \infty} p(v_k) \leq p(0).
\end{align*}
Therefore
\begin{align*}
p(v_k) \to p(0).
\end{align*}
Since $(v_k)$ was arbitrary, this is exactly stability of $p$ at $0$ relative to $V \cap \operatorname{dom} p$.
[/step]
