On proving a property of Bouligand (tangent) cones

Definition. The Bouligand (tangent) cone for the set $$X \subset \mathbb R^n$$ in the point $$x \in X$$ is defined as follows:

$$\mathcal B_X(x) = \{ d \in \mathbb R^n \, \colon \, \exists \{t_k\} \subset \mathbb R^+, t_k \to 0^+, \exists \{d_k\} \subset \mathbb R^n, d_k \to d \, \colon \, x + t_kd_k \in X, \forall k \in \mathbb N \}.$$

Property. Let $$C \subseteq \mathbb R^n$$ and $$D \subseteq \mathbb R^n.$$ Show that

$$\mathcal B_{C \cap D} (x) \subset \mathcal B_C(x) \bigcap \mathcal B_D(x).$$

My attempt. Consider an element $$d \in \mathcal B_{C \cap D}(x)$$ arbitrarily. Then, we can guarantee that there exists sequences $$\{t_k\}\subset\mathbb R^+$$, $$\{d_k\} \subset \mathbb R^n$$ such that $$t_k \to 0^+, d_k \to d$$ and $$x + t_kd_k \in C \cap D,$$ for every $$k \in \mathbb N.$$ Hence, we have that $$x + t_kd_k \in C$$ and $$x + t_kd_k \in D,$$ for every $$n \in \mathbb N.$$ This is obviously enough to show that $$d \in \mathcal B_C(x)$$ and $$d \in \mathcal B_D(x).$$ Therefore, $$d \in \mathcal B_C(x) \bigcap \mathcal B_D(x).$$ Since $$d$$ was arbitrary, it follows that $$\mathcal B_{C \cap D}(x) \subseteq \mathcal B_C(x) \bigcap \mathcal B_D(x).$$

My question. I am having issues proving that $$\mathcal B_{C \cap D}(x)$$ is STRICTLY contained in $$\mathcal B_C(x) \bigcap \mathcal B_D(x)$$. I understand showing a counterexample is enough, but I am having quite some trouble figuring one out. Can you give a counterexample?

Property. Let $$C \subseteq \mathbb R^n$$ and $$D \subseteq \mathbb R^n.$$ Show that $$\mathcal B_{C \cap D} (x) \subset \mathcal B_C(x) \bigcap \mathcal B_D(x).$$

Let "$$\subset$$" mean "is a subset of". Your attempt have proved the property correctly.

It is not always true that "$$\mathcal B_{C \cap D}(x)$$ is STRICTLY contained in $$\mathcal B_C(x) \bigcap \mathcal B_D(x)$$." It can happen that $$\mathcal B_{C \cap D} (x)$$ $$= \mathcal B_C(x) \bigcap \mathcal B_D(x).$$

For example, when $$C=D$$, we have
$$C \cap D=C=D$$,
$$\mathcal B_{C \cap D} (x)$$ $$=\mathcal B_C(x)$$ $$=\mathcal B_D(x)$$ $$=\mathcal B_C(x) \bigcap \mathcal B_D(x)$$.

For another example, let $$n=1$$, $$x=0$$, $$C=[-2,1]$$, $$D=[-1,2]$$.
Then $$C\cap D=[-1,1]$$.
$$\mathcal B_{C \cap D} (x)$$ $$=\mathcal B_C(x)$$ $$=\mathcal B_D(x)$$ $$=\mathcal B_C(x) \bigcap \mathcal B_D(x)$$ $$=\Bbb R$$.

It can also happen that $$\mathcal B_{C \cap D} (x)$$ $$\subsetneq \mathcal B_C(x) \bigcap \mathcal B_D(x)$$, of course.
For example, let $$n=1$$, $$x=0$$, $$C=\{0, \frac12, \frac14, \frac16, \cdots\}$$, $$D=\{0,\frac11,\frac13,\frac15,\cdots\}$$.
Then $$C\cap D=\{0\}$$.
$$\mathcal B_{C \cap D} (x)=\{0\}$$.
$$\Bbb R^{\ge0}$$ $$=\mathcal B_C(x)$$ $$=\mathcal B_D(x)$$ $$=\mathcal B_C(x) \bigcap \mathcal B_D(x)$$.

The meaning of "$$\subset$$" is not consistent across math literature. It might mean "is a proper subset of". It might mean "is a subset of", which includes the possibility of "is the same as". Only the latter meaning would make the property quoted above correct.

In order to avoid inconsistency and confusion, use either $$\subseteq$$ or $$\subsetneq$$, which are always clear.