Let $\Omega \subset \mathbb{R}^n$ open, $f \in C(\Omega)$ and $g \in C^1(\Omega)$ and assume that $f - g$ has a strict global minimum or maximum at $\overline{x} \in \Omega$.
What conditions on $f$ or $g$ make the following statement hold true?
There exists a sequence of functions $\{f_n\}$ such that $f_n \to f$ uniformly $\implies$ there is a sequence of points $x_n \to \overline{x}$ such that $f_n(x_n) \to f(x)$ and $f_n - g$ has a global minimum or maximum at $x_n$.
Note that $g$ plays no role in anything so I will omit it. The result holds, with the added caveat that there is possibly some finite subset of the $\{f_n\}$ with no minimum, under the following conditions:
Since $\partial\Omega$ is compact, there exists $\delta>0$ such that $f(\overline x)+\delta<\liminf\limits_{x\to a}f(x)$ for all $a\in\partial\Omega$. Let $N$ be such that $|f_n(x)-f(x)|<\frac\delta3$ for all $n\ge N$. We have
$$f_n(\overline x)<f(\overline x)+\tfrac\delta3<\liminf_{x\to a}f(x)-\tfrac{2\delta}3\le\liminf_{x\to a}f(x)-\tfrac\delta3$$
for any $a\in\partial\Omega$, and so in particular $f_n$ has a minimum in $\Omega$ for all $n\ge N$. Suppose $x_n$ minimizes $f_n$. Then $$f(x_n)=[f(x_n)-f_n(x_n)]+f_n(x_n)\le[f(x_n)-f_n(x_n)]+f_n(\overline x)\to f(\overline x).$$ If $x_{n_k}\to x_0\in\Omega$, this implies $f(x_0)\le f(\overline x)$ and hence $x_0=\overline x$ by uniqueness of the minimizer. On the other hand, if $x_{n_k}\to x_0\in\partial\Omega$ then
$$\liminf_{x\to x_0}f(x)\le\limsup_{k\to\infty}f(x_{n_k})\le f(\overline x)<\liminf_{x\to x_0}f(x)+\tfrac\delta3,$$
a contradiction. Hence the only limit point of the (bounded) sequence $\{x_n\}$ is $\overline x$, thus $x_n\to\overline x$. By uniform convergence we conclude $f_n(x_n)\to f(\overline x)$.
Note that this proof is inspired by $\Gamma$-convergence, and fairly closely follows the proof that if a sequence of functionals $\{F_n\}$ $\Gamma$-converges to $F$, then the minimum of $F_n$ converges to the minimum of $F$ (provided they all exist).