So, here's what I'm trying to prove:
Every orthonormal system is linearly independent.
Note:
I am aware of other questions of a similar nature. I'm trying to prove the theorem on my own so I'm doing my hardest to not look at those questions until someone verifies what I have done.
Proof Attempt:
Let $(V,\langle , \rangle)$ be a Euclidean Vector Space. Let $(v_1,v_2,\ldots,v_n)$ be an orthonormal system in that vector space. Now, consider the following sum:
$$\sum_{k=1}^{n} \alpha_k v_k = 0$$
We need to show that all of the real coefficients are, in fact, going to be 0. Let $k \in \{1,2,\ldots,n\}$ be arbitrary but fixed. Then, consider the following:
$$\alpha_k ||v_k||^2 = \alpha_k = \sum_{i=1}^{n} \langle v_k , \alpha_i v_i \rangle$$
The bilinearity of the inner product gives us the following:
$$\alpha_k = \langle v_k, \sum_{i=1}^{n} \alpha_i v_i \rangle = \langle v_k , 0 \rangle = 0$$
Since our chosen $k$ was arbitrary, it follows that:
$$\forall k \in \{1,2,\ldots,n\}: \alpha_k = 0$$
That is exactly what we wanted for linear independence. Hence, this proves the desired result.
Does the proof above work? If it doesn't, how can I fix it?
Your work is fine. A more direct approach is $$0 = \left\|\sum_{k=1}^n \alpha_k v_k \right\|^2 = \sum_{k=1}^n \sum_{\ell=1}^n \alpha_k \alpha_\ell \langle v_k, v_\ell \rangle = \sum_{k=1}^n \alpha_k^2.$$