Normal cone to the union of sets

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For a set $A \subseteq \Bbb R^n$ and a point $\bar{x} \in A,$ the limiting (Mordukhovich) normal cone of $A$ at $\bar{x}$ is defined as

$$N(\bar{x},A):= \limsup_{x\to \bar{x}} \widehat{N}(x,A),$$ where

$$\widehat{N}(x, A):= \left\{u \in \Bbb R^n \mid \limsup_{x'\to x,\; x'\in A}\frac{u^\top (x'-x)}{\|x'-x\|}\leq 0\right\} $$ is the so-called Frechet normal cone and the limit is understood in the sense of Painleve-Kuratowski. Furthermore, for a functional $f: \Bbb R^n \to \Bbb R$, the limiting subdifferential of $f$ at $\bar{x}$ is defined as

$$\partial f(\bar{x}):= {u \in \Bbb R^n \mid (u,-1) \in N((\bar{x},f(\bar{x}), epi f) },$$ where $epi f$ is the epigraph of $f.$ All of these definitions and more are from the book "Variational Analysis and Generalized Differentiation" by Boris Mordukhovich.

Now, consider two closed sets $A,B \subseteq \Bbb R^n$ and a point $\bar{x} \in A\cap B.$ My question is: does the inclusion

$$N(\bar{x}, A\cup B) \subseteq N(\bar{x}, A) \cup N(\bar{x},B)$$ holds? If so, does there exists a reference for the result or a tighter upper bound?

This question seems to be very natural, given the normal cone intersection formula, see Theorem 3.4 in Mordukhovich's book. However, I am not able to find such a result in the literature, and I need to cite it. My attempt at a proof is the following:

Consider $\delta_A$ and $\delta_B,$ the indicator functions of $A$ and $B$ respectively. Then, it is easy to verify that $\delta_{A\cup B}= \min\{\delta_A, \delta_B\}.$ By Proposition 1.79 in Mordukhovich's book, we have $\partial \delta_A(\bar{x}) = N(\bar{x},A).$ Therefore,

$$N(\bar{x}, A\cup B) = \partial (\min\{\delta_A, \delta_B\})(\bar{x}).$$ Moreover, by Proposition 1.113 (Mordukhovich's book) we have $$\partial (\min\{\delta_A, \delta_B\})(\bar{x}) \subseteq \partial \delta_A(\bar{x}) \cup \partial \delta_B(\bar{x}) = N(\bar{x}, A) \cup N(\bar{x},B),$$ and hence the upper bound follows.

Is this correct? A direct reference would be more beneficial in any case.