If $S$ contains uniformly lipschitz functions (As it holds for the sets $S$ and $M$ you mentioned in additional part of your question) Then:
If $h$ monotonically approaches to $h$ then $\partial h$ approaches to $\partial f$ (in terms of the following distance ) $$d(\partial h , \partial f) :=
\underset{ x \in X}{\sup} \underset{ v \in \partial f(x), w \in \partial h(x)}{\inf}||v-w||_2
$$
In another words, $$h_n \uparrow f \quad \quad \Rightarrow \quad \quad \partial h_n \to \partial f $$
Proof Assume $h_n \uparrow f $. Since $h_n$ are uniformly Lipschitz then there exist a $M > 0$ such that $$\| h_n(x) - h_m(y)\| \leq M \| x -y\|$$ this implies for all $x \in X$, $n \in N$, and $ v \in \partial h_n (x)$ we have $\| v \| \leq M.$
Now pick $x , y \in X $ and fix $x \in X$, assume $\epsilon > 0$ is arbitrary. Since $h_n \uparrow f $, large enough $n$ we have $$ h_n(y) \leq f(y), \quad f(x) -\epsilon \leq h_n (x) $$
Pick $v_n \in \partial h_n(x)$ therefore:
$$ \langle v_n , y- x\rangle \leq h_n(y) - h_n(x) \leq f(y) - f(x) + \epsilon . $$
Since $\{ v_n \}$ is bounded sequence, without loss of generality we may assume $v_n \to v .$ So by letting $ n \to \infty$ in above we get
$$ \langle v , y- x\rangle \leq f(y) - f(x) + \epsilon $$
Now because $\epsilon > 0$ was arbitrary, let $\epsilon \to 0$ in above so we arrive $$ \langle v , y- x\rangle \leq f(y) - f(x) $$ which means $ v \in \partial f(x)$
Presume $f(x),h_i(x)$ and $g_j(x)$ are continuously differential function, then
A constraint qualification is necessary to guarantee that local minimum implies KKT point. I.e., constraint qualification is required to make KKT conditions necessary for minimum.
If one or more of $f(x),h_i(x)$ and $g_j(x)$ are non-convex, then (in general) strong duality does not necessarily hold. That's what makes non-convex optimization difficult.
See https://en.wikipedia.org/wiki/Strong_duality
Best Answer
$$x^* \text{ minimizes } f(x) \Longleftrightarrow 0\in\partial f(x^*)$$ is trivial by definition of subgradients: $$f(x) - f(x^*) \ge \partial f(x^*)^T (x-x^*)\quad \forall x.$$
Thanks to Michael Grant for his comments.