Prove that if $S\equiv\{v_1,...,v_n\}$ is linearly independent, then every subset of $S$ is also linearly independent I need help with the proof of this statement:
If $S=\big\{ \vec{v_1},...,\vec{v_{n}} \big\}$ is a set of linearly independent vectors then any non-empty subset of $S$ contains only linearly independent vectors.
I tried using the definition
$\vec{v_1},...,\vec{v_n}$ are l.i. $\iff \lambda_1\vec{v_1}+...+\lambda_n\vec{v_n}=\vec{o} \implies \lambda_1=...=\lambda_n=0$
But is this enough to conclude that, for example, considering a subset $S'=\big\{ \vec{v_1},...,\vec{v_{n-1}} \big\}$ for sure $\lambda_1\vec{v_1}+...+\lambda_n\vec{v_{n-1}}=\vec{o} \implies \lambda_1=...=\lambda_{n-1}=0$ ?
It makes sense of course but is it enough for the proof?
Thanks for your help

There is a theorem that states

If a set of vectors $S$ is linearly independent, then any subset of $S$ is also a linearly independent set of vectors.

Proof
Let T be a subset of S. Let us reorder the vectors of S in a way to help us with the proof.
Make $T = \big \{ v_1, \dots ,v_k \big \}$ and $S = \big \{ v_1, \dots ,v_k, \dots , v_n \big \}$
Let's look at the equation
$$c_1v_1 + c_2v_2 + \dots +  c_kv_k = 0 $$
Make $c_{k+1} =c_{k+2} = \dots = c_n = 0$ then we can rewrite the initial equation as
$$c_1v_1 + c_2v_2 + \dots +  c_kv_k + 0v_{k+1} + \dots + 0v_{n} = 0 $$
Since S is linearly independent, then this forces $c_1 = c_2 = \dots = ck = 0$.
$\implies$ T is linearly independent.
