$\textbf{Question}$: Show that there exist an uncountable subset $X$ of $\mathbb{R}^{n}$ with property that every subset of $X$ with $n$ elements is a basis of $\mathbb{R}^{n}$.
$\textbf{My Attempt}$: For $n= 2$, if we take $X = \{(\cos \theta, \sin \theta): 0<\theta < \frac{\pi}{2}\}$, then $X$ has such property. But for $n\geq 3$, I can not find any way. Welcome for answer of this question.
You can consider all such subsets of $\mathbb R^n$ (not necessarily uncountable). They form a poset under inclusion and it is easy to see that every chain has an upper bound (their union). By Zorn's lemma, there exists a maximal subset $S$ of $\mathbb R^n$ with the required property. If $S$ were countable, then $\mathbb R^n$ would be a countable union of hyperplanes, and this is not true: Show that $\mathbb{R}^n$ cannot be written as a countable union of proper subspaces
The set of vectors $(1, x, x^2, \ldots, x^{n-1})$ with $x > 0$ works, because the condition amounts to the fact that a Vandermonde determinant is nonzero.