S is a linearly dependent set with infinitely many elements... Is it possible, that every finite subset of $S$ is linearly independent?
If $S$ is the standard basis for the set of all sequences of real numbers along with the Fibonacci sequence... Then S is dependent... But every finite subset of S will be independent.
Something feels wrong about this.
 A: Linear independence is by definition a property that has finite character.  I.e. a set of vectors is linearly independent iff every one of its finite subsets is linearly independent.
A set $S$ is linearly independent precisely if there is no finite subset of $S$ such that $0$ is a non-zero linear combination of them.  That's the definition.
In some contexts, especially, when working with Hilbert spaces, one uses infinite linear combinations, but not for defining the concept of linear independence.  These infinite linear combinations are defined using a notion of convergence of series that comes from the norm on the space.  Then one might speak of a diffent kind of independence, where $S\cup\{v\}$ is not "independent" since $v$ is an infinite linear combination of members of $S$.  Equivalently, $v$ is in the closure of the span of $S$.  But that's not "linear dependence" as that term is normally presumed to be defined.
Your example involving Fibonacci numbers is by this definition a linearly independent set, since the sequence of Fibonacci numbers is not a finite linear combination of the members of the standard basis.  In contexts where infinite linear combinations are dealt with, this only means linear dependence and independence by the standard definitions may not be the concepts you want to use.
A: We say a set $S$ is linearly dependent if there is some finite nontrivial (i.e. not all coefficients are $0$) linear combination of vectors in $S$ which is equal to $0$. Thus if an infinite set $S$ is linearly dependent, the set of vectors in said finite linear combination is a finite linearly dependent subset. So no, not every finite subset can be linearly independent.
