I see that questions relating to this lemma have been asked here and here .
But my question is quite different in the sense , that i believe that i have understood what the author intends to say but i need a confirmation if my understanding is correct .
The lemma is : 
So this is my understanding : To prove the lemma , we would prove that $$\exists c> 0 ,\text{ for which} \forall \alpha_{i} \in R , \text{ we have }||\alpha_{1}x_{1} +\alpha_{2}x_{2} +\alpha_{3}x_{3} +.....+ \alpha_{n}x_{n}|| \geq c(|\alpha_{1}|+|\alpha_{2}|+|\alpha_{3|}+.....+|\alpha_{n}|) \tag{1} $$ where the set of vectors $$\{x_{1},x_{2},.....,x_{n}\}$$ are independent and $x_{i} \in X$ where $X$ is a normed space with some norm.
Now the first idea that comes to me is that because the set of vectors are independent .We know , from linear algebra , that $$\alpha_{1}x_{1} +\alpha_{2}x_{2} +\alpha_{3}x_{3} +.....+ \alpha_{n}x_{n} = 0 $$ only when all $\alpha_{i}$ are zero , in which case the inequality of the lemma will hold true for all $c$ . But let $\alpha_{i} \neq 0$ . In this case the quantity $$\alpha_{1}x_{1} +\alpha_{2}x_{2} +\alpha_{3}x_{3} +.....+ \alpha_{n}x_{n} \tag{1}$$ will have some value but we don't know what it is but the norm over it would be some positive value .
$\textbf{(1) }$So intuitively , we can think that there exists and infinitesimally small $c \neq 0 $ and this proves the lemma. Is this reasoning correct ?
$\textbf{(2) }$Now , separately i had a look at author's explanation . And this is what i think he is saying . After some initial algebraic manipulation , we arrive to prove that : $$ \exists c \gt 0 \text{ such that } ||\beta_{1}x_{1}+\beta_{2}x_{2}+\beta_{3}x_{3}+....+\beta_{n}x_{n}|| \geq c \tag{2}$$ where $$\sum_{i=1}^{i=n}|\beta_{i}| = 1$$ . Now to prove $\textit{ equation 2 }$this , what the author is trying to do is that he creates a sequence which contains all the possible values for $$\beta_{1}x_{1}+\beta_{2}x_{2}+\beta_{3}x_{3}+....+\beta_{n}x_{n} \tag {3} $$ with a norm over it .And then he proves that this sequence converges to a a point , and the value of the point is not $0$ , and hence all elements of the of the sequences are not $0$ and hence the same intution , that followed in $\textit{Point (1.) }$follows here . So is this deduction by me correct ? If that is the case , why did not the author just used $\textit{Point (1.) }$ ?
$\textbf{(3) }$Lastly i want to confirm my understanding of how he created the sequence : Let the sequence be denoted by $(y_{m})$ with the general element given by $$y_{m}= \beta_{1}^{(m)}x_{1}+\beta_{2}^{(m)}x_{2}+\beta_{3}^{(m)}x_{3}+....\beta_{n}^{(m)}x_{n}$$ . So that for m =1 ,2 , we have : $$y_{1} =\beta_{1}^{(1)}x_{1}+\beta_{2}^{(1)}x_{2}+\beta_{3}^{(1)}x_{3}+....\beta_{n}^{(1)}x_{n} \\ y_{2} =\beta_{1}^{(2)}x_{1}+\beta_{2}^{(2)}x_{2}+\beta_{3}^{(2)}x_{3}+....\beta_{n}^{(2)}x_{n} \\ .\\ .\\ .\\ y_{m} =\beta_{1}^{(m)}x_{1}+\beta_{2}^{(m)}x_{2}+\beta_{3}^{(m)}x_{3}+....\beta_{n}^{(m)}x_{n} \tag{4} $$ Now we know that $\sum_{i=1}^{i=n}|\beta_{i}^{m}| =1 $ , this implies that $|\beta_{i}^{m}| \leq 1$ . So we can say that the sequence $$(\beta_{1}^{1},\beta_{1}^{2},\beta_{1}^{3}...)$$is bounded and by the BW theorem , there exists its sub sequence which is convergent and let that point of convergence be called $\beta_{1}$ .The author then writes $\textit{Let } (y_{1,m}) \textit{ be the corresponding sub sequence of } y_{m}$. I want to confirm if the general element of the $(y_{1,m})$ looks like : $$y_{1,m} = \beta_{1}x_{1}+ \beta_{2}^{m}x_{2}+\beta_{3}^{m}x_{3} +.....+ \beta_{n}^{m}x_{n}$$ Is that correct ?
$\textbf{(4) }$ And we repeat the process and reach the $\textit{ expression 3 }$.
Edit 1 : I wrongly ordered the quantifiers as pointed out by @Matthew Towers. It has been fixed now. Also the language around c has been changed to make "c" look more like a constant. Edit 2 : Typo around $|.|$ in equation 1 fixed.