There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent.

Try $\left(\begin{matrix}1\\t\\t^2\end{matrix}\right)$ with $t\in\mathbb R$. Do you know to compute $$\left\vert\begin{matrix}1&1&1\\r&s&t\\r^2&s^2&t^2\end{matrix}\right\vert? $$

Three vectors in $\mathbb{R}^3$ are linearly dependent if and only if they lie in a plane.

Consider the following process for building $S$. We can start with the empty set, and choose any two vectors $v_1, v_2 \in \mathbb{R}^3$ and add them to $S$. Then to choose a third vector $v_3$ to add to $S$, we must make sure it is not in the unique plane containing (i.e. spanned by) $v_1$ and $v_2$. Thus $v_3$ can be any vector in $\mathbb{R}^3 \backslash span(v_1, v_2)$.

Similarly, if at some stage $S = \{v_1, \ldots, v_k\}$, we can add to $S$ any vector $v_{k+1}$ in $\mathbb{R}^3 \backslash \bigcup_{x_i, x_j} span(x_i, x_j)$. Note that $\bigcup_{x_i, x_j} span(x_i, x_j)$ is a finite union of planes, so it can never be all of $\mathbb{R}^3$. In this way we can choose an infinite set with the desired property.

Consider vectors of the form $v_x=(1,x,x^2)^T$. Then for any distinct $x,y,z\in \mathbb R$ matrix $(v_x v_yv_z)$ is nonsingular (Vandermonde matrix), so $v_x$, $v_y$, $v_z$ are linearly independent.

The parametric curve $t\mapsto(1,t,t^2)$ has the property that any three distinct points on it are linearly independent (without the "distinct" one clearly cannot have a solution). To check, just compute the determinant, which is Vandermonde.

Inspired by the Vandermonde matrix example, let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a strictly convex function. Then $$\left\vert\begin{matrix}1&1&1\\x&y&z\\f(x)&f(y)&f(z)\end{matrix}\right\vert\not=0$$ for any three distinct numbers $x,y,z$.