Let $K\ (\subseteq \mathbb{R}^n)$ be a closed convex cone and $\{K^r\}_{r=0}^\infty$ be a family of closed convex cones satisfying $K^r \subseteq K^{r+1} \subseteq K\ (\forall r)$.
Assume that ${\rm int}(K) \subseteq \bigcup_{r=0}^\infty K^r$ and there exists $x \in {\rm int}(K)$.
Then, does there exist $r_0$ such that $x \in {\rm int}(K^{r_0})$?
Besides, can we remove the assumption that $\{K^r\}$ are closed?
Intuitively, this proposition is true, but I cannot prove it rigorously. Note that a similar problem appears in the proof of Theorem 6.1 in the following paper: https://www.jstor.org/stable/25151657?seq=1#metadata_info_tab_contents
This is a bit of a round-about way of proving it, but it works. (There's probably a more straightforward way of doing this.)
First note that almost all of the cones $K_0,K_1,\dots$ must have non-empty interior. (Indeed, it is impossible for all of them to have empty interior, so there must exist an index $j\in\mathbb{N}$ such that $\operatorname{int}(K_r)\neq\emptyset$ for every $r\geq j$.)
Now let $x\in\operatorname{int}(K)$. We want to show that there exists an index $r_0\in\mathbb{N}$ for which $x\in\operatorname{int}(K_{r_0})$. By the assumption that $$\operatorname{int}(K)\subseteq\bigcup_{r\in\mathbb{N}} K_r,$$ there is an index $r_1\in\mathbb{N}$ for which $x\in K_{r_1}$. By the observation in the preceeding paragraph, we may suppose without loss of generality that $\operatorname{int}(K_{r_1})$ is non-empty. If $x\in\operatorname{int}(K_{r_1})$ then we are done, so suppose that $x\not\in\operatorname{int}(K_{r_1})$. (That is, $x$ is on the boundary of $K_{r_1}$.) There exists a vector $y\in\mathbb{R}^n$ such that $$x-y\in\operatorname{int}(K_{r_1}) \qquad\text{and}\qquad x+y\not\in K_{r_1}.$$ As $x\in\operatorname{int}(K)$, there exists a value $\epsilon>0$ small enough such that $x+\epsilon y\in\operatorname{int}(K)$. Define a new point $x_1=x+\epsilon y$. For this new point $x_1\in\operatorname{int}(K)$, there exists an index $r_0\in\mathbb{N}$ with $r_0>r_1$ such that $x_1\in K_{r_0}$. As it is the case that $$x-y\in\operatorname{int}(K_{r_1}) \qquad\text{and}\qquad K_{r_1}\subseteq K_{r_0},$$ it follows that $x-y\in\operatorname{int}(K_{r_0})$ as well. Every nontrivial convex combination of the points $x_1\in K_{r_0}$ and $x-y\in\operatorname{int}(K_{r_0})$ must be in $\operatorname{int}(K_{r_0})$. That is, for every $t\in(0,1)$, it holds that $$tx_1 + (1-t)(x-y)\in\operatorname{int}(K_{r_0}).$$ Choosing $t = \frac{1}{1+\epsilon}$ gives us the desired result that $x\in\operatorname{int}(K_{r_0})$.