I need to proof the following theorem and I would appreciate if you could check if my proof is sound. I'm generally new to proofs, so any recommendations on proof strategy/style are more than welcome. Thank you.
$\textbf{Theorem:}$
Let $S=\{\textbf{v}_1, \textbf{v}_2, ..., \textbf{v}_k\}$ be a linearly independent subset of some vector space $V$. Let $\textbf{u} \in V$. If $\textbf{u} \notin \operatorname{span}(S)$, then $S \cup \{\textbf{u}\}$ is linearly independent.
$\textbf{Proof:}$ I use a contrapositive proof, i.e. $\lnot Q \Rightarrow \lnot P $. Suppose $S \cup \{ \textbf{u} \}$ is linearly $\textbf{dependent}.$ Then, by definition of linear dependence, there exist $\alpha_1, \alpha_2, ..., \alpha_{k+1} \in \mathbb{R}$ not all zero s.t.
$$\sum_{i=1}^{k+1}\alpha_i v_i = 0$$ $$= \sum_{i=1}^{k}\alpha_i v_i + \alpha_j u = 0$$ where, without loss of generality, $\alpha_j u= \alpha_{k+1}v_{k+1}$.
Then it follows, $$\alpha_j u = -\sum_{i=1}^{k}\alpha_i v_i$$ $$= u =\sum_{i=1}^{k}\alpha_i^{'} v_i$$ where $\alpha_i^{'} = \frac{-\alpha_i}{\alpha_j}$, for $\alpha_j \neq 0$, implying that $u \in \operatorname{span}(S) \cup \{u\}$.
Thus, we can conclude that if $u \notin \operatorname{span}(S)$, $S \cup \{u\}$ is linearly independent. This completes the proof.
Your proof is essentially correct, but I would use more expressive notations, along the following lines:
Suppose$S\cup\{\mathbf{u}\}$ is linearly dependent. Then there exist coefficients $\alpha_1,\dots,\alpha_k\:$ and $\:\beta$, not all $0$, such that $$\sum_{i=1}^k\alpha_i\,\mathbf{v}_i+\beta\, \mathbf{u}=0.$$ Observe that $\beta\ne 0$, for otherwise, $S$ would be linearly dependent, contrary to the hypotheses. So we can deduce that $$\mathbf u=\sum_{i=1}^k\Bigl(-\tfrac{\alpha_i}\beta\Bigr)\mathbf{v}_i,$$ which means$\;\mathbf u\in\operatorname{span} S$.