 Solution

The Steinitz exchange lemma says that if $w_1,\dots ,w_m$ span a vector space and $v_1,\dots ,v_r$ are linearly independent, then $r\le m$, and after reordering the $w_j$, the family $v_1,\dots ,v_r,w_{r+1},\dots ,w_m$ still spans. Here is a proof. Suppose the first $k-1$ replacements have been made. Express $v_k$ in that spanning family. Some coefficient of a remaining $w_j$ must be nonzero, since otherwise $v_k$ belongs to the span of $v_1,\dots ,v_{k-1}$, contrary to linear independence. Solve for that $w_j$ and replace it by $v_k$; this preserves spanning. If $r>m$, the first $m$ replacements leave a spanning family $v_1,\dots ,v_m$, contradicting linear independence of the first $m+1$ vectors. This proves both assertions.

Now suppose $S$ is linearly independent and spans ${\mathbb{R}}^n$. Express each member of the standard basis using finitely many members of $S$, and let $S_0$ be the finite union of those members. Then $S_0$ spans. No element of $S\setminus S_0$ can exist, since it would be a linear combination of $S_0$ and violate linear independence. Thus $S$ is finite. Applying the Steinitz exchange lemma in both directions to $S$ and the standard basis gives $|S|\le n$ and $n\le |S|$. Hence $\boxed{|S|=n}$.