Here's a good one, which essentially shows that every closed $2$-dimensional subset of the plane contains a region of the plane.

Theorem. Let $X\subset \mathbb R^2$ be closed and $x\in X$. Suppose that for every $\varepsilon\in (0,1]$, $X$ contains an arc along the circle of radius $\varepsilon$ centered at $x$. Then $X$ contains a whole region of the plane.

Proof: Enumerate $\mathbb Q\cap [0,2\pi]$ as $q_1,q_2,q_3\ldots$. For each $\varepsilon\in (0,1]$ let $S_\varepsilon(x)$ be the circle of radius $\varepsilon$ centered at $x$. When $n\neq m$, let $S^{n,m}_\varepsilon(x)$ be the arc of points of $S_\varepsilon(x)$ at angles between $q_n$ and $q_m$ (in radians). Define $$A_{n,m}=\{\varepsilon\in [0,1]:S^{n,m}_\varepsilon(x)\subset X\}.$$ By assumption, $$X=\bigcup_{n,m\geq 1}A_{n,m}.$$ Note that each $A_{n,m}$ is closed in $(0,1]$, since $X$ is closed in $\mathbb R^2$. By BCT, there exist $n$ and $m$ such that $A_{n,m}$ contains an interval $I$ of $(0,1]$. Then $X$ contains the region consisting of all points $z$ such that the angle of $z$ relative to $x$ is between $q_n$ and $q_m$, and $d(x,z)\in I$.