The general statement "If $S$ is a spanning set, and $L$ is linearly independent then $|L|\leq|S|$" is unfamiliar to me, and I'm not sure it was investigated enough to merit an answer either. It does imply that every two bases have the same cardinality, so it does require some weak form of the axiom of choice. My guess would be that it requires a bit more than just every two bases have the same cardinality.
However, it seems that you are interested in the particular case of a countable basis. That is, if there is a countable basis, does every linearly independent set inject into a spanning set? Well, we can do better. We can prove that every linearly independent is countable.
Claim. If $V$ is a vector space over $K$, and $B$ is a countable basis for $V$, then every linearly independent set have size $\leq\aleph_0$.
If $B$ Is finite, then we know it to be true, and it is not the interesting case anyway.
Fix an enumeration of $B=\{b_n\mid n\in\Bbb N\}$. And let $L$ be a linearly independent set, and assume by contradiction that $|L|\nleq\aleph_0$. For every $\ell\in L$ take $\langle\ell_n\mid n\in\Bbb N\rangle$ to be the sequence of coefficients for the unique sum of $\ell$ in term of $b_n$'s. Note that all but finitely many are zero.
Clearly $\ell\mapsto\langle\ell_n\mid n\in\Bbb N\rangle$ is injective. Since $L$ is not countable, there is an uncountable subset of elements which have exactly the same non-zero indices, so without loss of generality it will be all of $L$, and without loss of generality those indices are $0,\ldots,n-1$.
Therefore $L$ itself is linearly independent as a subset of $W=\operatorname{span}\{b_i\mid i<n\}$. But this is a contradiction since $W$ has a finite dimension, and therefore every linearly independent set is finite, but $L$ is uncountable. $\quad\square$
You need to use the fact that every linearly independent set can be extended to a basis.
Then you can take any nonzero vector $v$, extend $\{v\}$ to a basis $B$, and consider the span of $B\setminus\{v\}$. Being finitely dimensional, it means that $B\setminus\{v\}$ is finite, so $B$ is finite, and so $V$ has a finite dimension.
It might be relevant to point out that the fact "every linearly independent set can be extended to a basis" is equivalent to the axiom of choice. Of course we need only a small fraction of choice for this specific proof (although that would alter the formulation a bit).
Nevertheless, it is consistent without the axiom of choice that there is a vector space which is not finitely dimensional, but every proper subspace does in fact have a finite dimension. Weird.
Best Answer
Assuming the axiom of choice, yes. If $W$ is a subspace of $V$, choose a basis of $W$, and extend it to a basis of $V$; now recall that the cardinality of a basis is unique, even for infinite-dimensional spaces, and that is enough to finish the argument.
Without the axiom of choice, however, there is no real notion of dimension anymore. It is possible for a space to have two bases of different cardinalities, or a space with a basis will have a subspace without a basis, etc.