Given an infinite bounded set A in $R^n$,$2\leq n$, show there are infinite boundary points I started studying multivariable caclulus, and am having problems with this exercise:

Given an infinite bounded set A in $R^n$,$2\leq n$, show there are infinite boundary points

Attempt: Going from the assumption that A has finite boundary points, we can assume it has an interior point. I believe it's possible to construct a ball around that point with large enough radius (assuming A is bounded by M, 3M should suffice), and prove that there's a boundary point on every line between a boundary point of that ball and the interior point we selected. This ended up being very complex and I didn't manage to work out all the details nicely, so I figure there must be something easier!
Thanks!
 A: For the $n=2$ case:
Let $A$ be such a set and let $L_x$ denote the vertical line that passes through $(x,0)$. Note that any boundary point of $L_x\cap A$ (as a subset of $L_x$) is a boundary point of $A$, as if an open ball $B$ contains a point in $L_x$ outside $A\cap L_x$ then this point lies outside $A$ as well. Since $A$ is bounded, for each $x\in\mathbb R$ either $L_x\cap A$ is empty or it has at least one boundary point (this is not hard to show). If there are infinitely many $x\in \mathbb R$ such that $L_x\cap A$ is nonempty, then we have infinitely many distinct boundary points and are done. Otherwise, for each $x$ such that $L_x\cap A$ is nonempty we have some $\epsilon>0$ such that $L_y\cap A$ is empty for any $y$ such that $|x-y|<\epsilon$. Thus any point $p\in L_x\cap A$ is such that $B_r(p)$ contains $B_{\min\{r,\epsilon\}}(p)\setminus L_x$ which is nonempty and does not intersect $A$, so is a boundary point of $A$. Hence all points in $A$ are boundary points, so $A$ has infinitely many boundary points.
For $n>2$, the argument is very similar. I will let you work out the details.
A: If $A$ has no interior point then every point of $A$ is on the boundary so there are infinitely many boundary points. Otherwise $A$ contains two points such that the line $L$ which connects them is contained in $A$. Let $v$ be a nonzero vector that is orthogonal to $L$ (this exists since $n \geq 2$). For every point $x \in L$ consider the set $ \{\lambda \geq 0: x+\lambda  v \in A\}$. It is non-empty because it contains $0$ and it is bounded because $A$ is bounded. Let $\lambda_0$ the supremum of this set. Then $x+\lambda_0 v$ is a boundary point of $A$. The lines $x + \mathbb Rv$ ($x \in L$) are disjoint so $A$ has infinitely many boundary points.
