Let $ S = \{(1, 0, 0), (0, 1, 0)\} $. Then, $ S $ is linearly independent, as is easily seen, on the other hand $ (0, 0, 1) \notin \textrm{span}\, S $, therefore $ S $ does not span $ \mathbb{R}^3 $.
On the other hand, your intuition is partly correct due to the following result:
Theorem. Let $ V $ be a vector space, $ L $ a linearly independent subset and $ S $ a subset that spans $ V $. Then, $ |L| \leq |S| $.
Proof. Let $ S = S_0 = \{ s_i : 1 \leq i \leq n \} $ and let $ L = \{ b_i : 1 \leq i \leq m \} $. We construct a sequence of spanning sets. Given $ S_k $, construct the set $ S_{k+1} $ as follows: $ S_k $ is a spanning set, therefore we may write $ b_k = \sum c_i {s_k}_i $ where $ {s_k}_i \in S_k $ and the $ c_i $ are members of the field of scalars. As $ L $ is linearly independent, there must be a vector on the right hand side which is not an element of $ L $, let one such vector be ${s_k}_j$. Therefore, removing $ {s_k}_j $ from the set $S_k$ and replacing it with $ b_k $ gives us a spanning set with the same number of elements as $ S_k $ (as ${s_k}_j$ can be expressed as a linear combination of the elements of this new set). Define this set to be $ S_{k+1} $.
With this construction, a new element of $ L $ is added to the sets $ S_i $ at each step, however the cardinality of the sets remains unchanged. The construction halts at $ S_m $, which contains all elements of $ L $, therefore $ L \subseteq S_m $ and $|L| \leq |S_m| = |S|$, which establishes the result.
Corollary. Let $ V $ have dimension $ n $ over its field of scalars and let $ L $ be a linearly independent subset of $ V $ which has $ n $ elements. Then, $ L $ is a basis of $ V $.
Proof. Let $ B $ be a basis for $ V $, then $ |B| = n $. Consider the set $ L' = L \cup \{v\} $ for any $ v \in V $ and $ v \notin L $. This set has $ n+1 $ elements. However, any linearly independent subset of $ V $ can have at most $ n $ elements by the above theorem, as $ B $ is a spanning subset. Therefore, $ L' $ is linearly dependent, and in particular $ v $ can be expressed as a linear combination of the elements of $ L $ (otherwise $ L $ would be linearly dependent), which establishes that $\textrm{span}\, L = V $. By definition of a basis, $ L $ is a basis of $ V $.
Therefore, if your linearly independent subset has as many elements as the dimension of your vector space. then it has to span your space.
Suppose the vectors $v_1,\dots,v_k$ are linearly dependent.
Then there exist non-trivial (not all zero) coefficients $a_1,\dots,a_k$ so that
$$
a_1v_1+\dots+a_kv_k=0.
$$
Now express this in the basis $B$:
$$
0
=
[a_1v_1+\dots+a_kv_k]_B
=
a_1[v_1]_B+\dots+a_k[v_k]_B.
$$
Therefore the vectors $[v_1]_B,\dots,[v_k]_B$ are linearly dependent.
If you assume that the vectors $[v_1]_B,\dots,[v_k]_B$ are linearly dependent, you can follow the same steps backwards to show that $v_1,\dots,v_k$ are linearly dependent.
We have shown that one set of vectors is linearly dependent if and only if the other one is.
Therefore one set of vectors is linearly independent if and only if the other one is.
Best Answer
If $k>n$, then the basis is not linearly independent, that is, there exists an $x$ such that $v_x$ can be expressed as a linear combination of other elements of $S$.
If $k<n$, then the basis does not span the space. That is, there exists a $w \in \mathbb{R}^n$ that cannot be expressed as linear combination of elements of $S$.