 Solution

For the equal-cardinality basis criterion, because $S$ spans $V$, repeatedly remove a vector whenever it is a linear combination of the others. Such a removal does not change the span. The finite process ends with a linearly independent spanning subset $T$, hence a basis of $V$.

If the original $S$ were linearly dependent, at least one removal would occur and $|T|<n$. But a basis of an $n$-dimensional vector space has $n$ vectors. Equivalently, apply the allowed claim to the basis $T$ and any fixed $n$-vector basis of $V$: the latter would wrongly be linearly dependent. Thus no removal was possible. The original spanning set is linearly independent and therefore a basis.