A set of vectors is linearly dependent if and only if there is a nontrivial linear combination of the vectors in the set which is zero. Nontrivial here means that there is at least one nonzero coefficient in the linear combination.
It should be immediately obvious then that you can append ANY set of vectors, finite or infinite, even uncountable, to a linearly dependent set and the result is still a linearly dependent set. That's because once you have a nontrivial linear combination summing to zero, you can throw any additional vectors into that sum with all additional coefficients identically zero; this means that (1) the sum is still zero, and (2) the linear combination is still nontrivial.
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
The answer is yes the set is still linearly independent.
Assume that missing vector is $v_n$ and let $$\lambda _1v_1 + \lambda _2v_2+....+\lambda _{n-1}v_{n-1} =0$$
You need to show $$ \lambda _1 = \lambda _2 =...= \lambda _{n-1} =0$$
Consider the linear combination $$\lambda _1v_1 + \lambda _2v_2+....+\lambda _{n-1}v_{n-1} +0v_n=0$$
Since $$v_1,v_2,...,v_n$$ are linearly independent we have $$ \lambda _1 = \lambda _2 =...= \lambda _{n-1} = \lambda _n =0$$
Which implies $$ \lambda _1 = \lambda _2 =...= \lambda _{n-1} =0$$
Thus the new set is also linearly independent.