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.
Best Answer
You cannot use Gauss Jordan elimination, you will have to use the definition of linear independence: A set is linearly independent if it is the case that a linear combination of its vectors equals zero if and only if the coefficients of that linear combination are all zero.
Hint: Any linear combination of $\{v_2, v_3, v_4\}$ can be thought of as a linear combination of $\{v_1, v_2, v_3, v_4\}$ where the coefficient of $v_1$ is zero. For getting that $v_1$ is not in the span, prove this by contradiction, i.e., if $v_1$ is in the span then $\{v_1, v_2, v_3, v_4\}$ is not linearly independent.