We prove it by induction on $\dim_{k}V$. If $\dim_{k}V=1$, then the assertion is obvious. By the hypothesis of induction, there is a $\sqrt{\alpha}$-orthogonal basis $(e_{1},\ldots,e_{r-1})$ of $V^{\prime}:=ke^{\prime}_{1}+\cdots+ke^{\prime}_{r-1}$ with respect to $\|\raisebox{1.72218pt}{.}\|$ such that

$$ke_{1}+\cdots+ke_{i}=ke^{\prime}_{1}+\cdots+ke^{\prime}_{i}$$

for $i=1,\ldots,r-1$. Choose $v\in V\setminus V^{\prime}$. As

$$\mathrm{dist}(v,V^{\prime}):=\inf\{\|v-x\|:x\in V^{\prime}\}>0,$$

there is $y\in V^{\prime}$ such that $\|v-y\|\leq(\sqrt{\alpha})^{-1}\mathrm{dist}(v,V^{\prime})$. We set $e_{r}=v-y$. Clearly $(e_{1},\ldots,e_{r-1},e_{r})$ forms a basis of $V$. It is sufficient to see that

$$\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}+e_{r}\|\geq\alpha\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r-1}|\|e_{r-1}\|,\|e_{r}\|\}$$

for all $a_{1},\ldots,a_{r-1}\in k$. Indeed, as $\|e_{r}\|\leq(\sqrt{\alpha})^{-1}\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}+e_{r}\|$, we have

$$\alpha\|e_{r}\|\leq\sqrt{\alpha}\|e_{r}\|\leq\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}+e_{r}\|.$$

If $\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}\|\leq\|e_{r}\|$, then

	$\displaystyle\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}+e_{r}\|$	$\displaystyle\geq\sqrt{\alpha}\|e_{r}\|\geq\sqrt{\alpha}\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}\|$
		$\displaystyle\geq\sqrt{\alpha}\left(\sqrt{\alpha}\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r-1}|\|e_{r-1}\|\}\right)$
		$\displaystyle=\alpha\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r-1}|\|e_{r-1}\|\}.$

Otherwise,

	$\displaystyle\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}+e_{r}\|$	$\displaystyle=\|a_{1}e_{1}+\cdots+a_{r-1}e_{r-1}\|$
		$\displaystyle\geq\sqrt{\alpha}\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r-1}|\|e_{r-1}\|\}$
		$\displaystyle\geq\alpha\max\{|a_{1}|\|e_{1}\|,\ldots,|a_{r-1}|\|e_{r-1}\|\},$

as required.

For the second assertion, it is sufficient to show the following lemma because it implies that the set $\{\|v-x\|\mid x\in V^{\prime}\}$ has the minimal value. ∎