Let be $V$ a vector space and $W\leq V$ a subspace of $V$ with basis $Y$. If we consider the quotient space $V/W$ , by Zorn's lemma, we can obtain a basis of $V/W$, denoted $\overline{S}$.
If $v\in V$, $\overline{v}=\beta_1\overline{\alpha_1}+\cdots+\beta_k\overline{\alpha_k}$ where $\alpha_i\in S$, then $v-(\beta_1\alpha_1+\cdots+\beta_k\alpha_k)\in W$ so $v=\beta_1\alpha_1+\cdots+\beta_k\alpha_k+\gamma_1\zeta_1+\cdots+\gamma_r\zeta_r$, where $\zeta_i\in Y$. Thus, $S$ (yes, without bar) and $Y$ generate $V$.
Finally, whit the same notation above, if $\{{\alpha_1,\cdots,\alpha_k}\}\subseteq S$ and $\{{\zeta_1,\cdots,\zeta_r}\}\subseteq Y$ the equation $$\beta_1\alpha_1+\cdots+\beta_k\alpha_k+\gamma_1\zeta_1+\cdots+\gamma_r\zeta_r=0$$ implies that each escalar is zero. Indeed, the equation implies that $$\beta_1\alpha_1+\cdots+\beta_k\alpha_k\in W,$$ then $\beta_1\overline{\alpha_1}+\cdots+\beta_k\overline{\alpha_k}=\overline{0}$ and it follows that each $\beta_i=0$ and by linear independence of $\{{\zeta_1,\cdots,\zeta_r}\}$ it follows that $\{{\alpha_1,\cdots,\alpha_k,\zeta_1,\cdots,\zeta_r}\}$ is linearly independent. Thus $Y$ can be extended to a basis of $V$.
PDT: I'm sorry, english is not my mother tongue.