Does every compact set in a normed space with a non-trivial interior has 2 path connected points in the boundary?

Let $k\subset\mathbb{R}^n$ be a compact space with $K^\circ \neq \emptyset$. Are there necessarily $a,b\in\partial K$ such that $$[a,b]=\{a+t(b-a)\mid t\in[0,1]\}\subset K$$ This is a lemma I want to prove in order to use mean-value type theorems in a compact set. I would appreciate notes regarding my own proof attempt (which isn't formal enough I believe) as well as other proofs.
Proof attempt:
There exists a point $x\in K^\circ$, so we can take $r=sup_r(\hat B _r(x)\subset K)$ and claim that $\partial K \cap\partial B_r(x)\neq\emptyset$. Now we can take a point $a$ from this intersection and look at the interval $$[a,\infty)=\{a-t(x-a)\mid t\in[0,\infty]\}$$ Now we can take another supremum: $$s=sup_\alpha \{a+t(x-a)\mid t\in[0,\alpha]\}\subset K$$ Since $K$ is compact in a normed space it is bounded, so $s<\infty$. It is also true that $s\geq2r$. Because of that $s$ is well defined. We define $b=a-s(x-a)\in\partial K$ and we have $$a,b\in\partial K \ \ \ s.t \ \ \ [a,b]\subset K$$

Your proof is fine and only your notation for supremum should be clarified: apply supremum/infimum simply on a set of real numbers.

First we take $$r\ :=\ \sup\{r: \hat{B_r}(x)\subseteq K\}$$

Then we take $$s\ :=\ \sup\{\alpha: [x,\,a+\alpha(x-a)]\subseteq K\}$$

Alternatively you can consider any point $a_0$ in $B_\varepsilon(x)\setminus\{x\}$ and do your second step twice, first for the ray $\overrightarrow{x,a_0}$, second for the ray $\overrightarrow{x,a'_0}$ where $a'_0=x-(a_0-x)=2x-a_0$.