Proof of linear independence of$~\boldsymbol{{x_1\over\Vert\boldsymbol x_1\Vert},x_2} ~$where$~\boldsymbol{x_1,x_2}~$are linearly independent. $$
\boldsymbol{x_1},~\ldots,\boldsymbol x_s:=n~\text{dimensional real vectors which are linearly independent}
$$
And it is obvious that there is no such$~\boldsymbol x_i=\boldsymbol 0~\text{for}~i\in\mathbb{N}_{s}^{*}~$since the$~s~$vectors are linearly independent.
$$
\boldsymbol a:={1\over\Vert\boldsymbol x_1\Vert}\boldsymbol x_1
$$
$$
\underbrace{\color{red}{\boldsymbol x_2~\textit{and}~\boldsymbol a~\textit{are linearly independent}~}}_{\text{Proposition which I want to derive. }}
$$
I know that a set as subset of$~\boldsymbol{x_1},~\ldots,\boldsymbol x_s~$is also linearly independent, from the quote of wikipedia,though I cannot prove it.
So $~\boldsymbol x_1~$and$~\boldsymbol x_2~$are linearly independent.
Which statement can be described as below.
$$
\underbrace{\mathrm{const}_1\cdot\boldsymbol x_1+\mathrm{const}_2\cdot\boldsymbol x_2=\boldsymbol 0}_{\text{This eqn can be held if only}~\mathrm{const}_1=\mathrm{const}_2=0~\text{is satisfied}}
$$
$$
\underbrace{ \color{blue}{\mathrm{const}_{a}\cdot {\boldsymbol x_1 \over \Vert\boldsymbol x_1\Vert  } +\mathrm{const}_{b}\cdot\boldsymbol x_2=\boldsymbol 0}  }_{\text{My brain has freezed from here} }
$$
How can I prove the red statement?
 A: Hint: Let const$_1=\frac{\text{const}_a}{\Vert x_1\Vert}$ and const$_2=$const$_b$.
I'm not sure that you fully understood what you're proving. So I'll elaborate.
Definition: vectors $x_1, \ldots, x_n$ are linearly independent if "if scalars $a_1, \ldots, a_n$ satisfies $a_1x_1+\ldots+a_nx_n=0$, then $a_1=\cdots=a_n=0$".
Claim: Given that $x_1,\ldots,x_n$ are linearly independent, you want to prove that $x_1$ and $cx_2$ are linearly independent, where $c\neq 0$.
Proof: A subset of a linearly independent set is also linearly independent. So $x_1, x_2$ are linearly independent. --- (*)
We want to prove the Claim, so suppose $b_1x_1 + b_2cx_2 = 0$ for some scalars $b_1,b_2$. --- (**)
But (*) implies that $b_1=b_2c=0$. (Read the definition carefully to see this. Note that (**) means that scalars $a_1:=b_1$ and $b_2:=b_2c$ satisfy $a_1x_1+a_2x_2=0$. These ":=" assignments are what I meant in the hint, but you need to understand how the proof works to apply it)
Since we assumed $c\neq 0$, we have $b_2=0$. To conclude, we have $b_1=b_2=0$. $\square$
A: Owing to the advice from @Benjamin Wang, I thought the following.
The key factor for me to this problem, is to think constants used in linear combination as variables. This expression may be illogical but I ignore it for the convenience.
$$
\underbrace{\color{fuchsia}{\mathrm{const}_{a}\cdot{\boldsymbol x_1\over\Vert\boldsymbol x_1\Vert}+\mathrm{const}_{b}\cdot\boldsymbol x_2=\boldsymbol 0}}_{\textit{Evaluation of}~\mathrm{const}_{a},~\mathrm{const}_{b}~\textit{s.t. holding above eqn is to be done}}
$$
$$
\mathrm{const}_{\gamma}:={\mathrm{const}_{a}\over\Vert\boldsymbol x_1\Vert}
$$
$$
\mathrm{const}_{\gamma}\cdot{\boldsymbol x_1}+\mathrm{const}_{b}\cdot\boldsymbol x_2=\boldsymbol 0
$$
As a necessary condition,$~\mathrm{const}_{\gamma}=0~$should be satisfed.
Hence$~\mathrm{const}_{a}=0~$should also be satisfied and this value is a sole one which$~\mathrm{const}_a~$can hold.
Now the top pink equation becomes
$$
\mathrm{const}_{b}\cdot\boldsymbol x_2=\boldsymbol 0
$$
Since$~\boldsymbol x_2\neq\boldsymbol0~$is held,$~\mathrm{const}_b=0~$should be satisfed.
$~\mathrm{const}_a=\mathrm{const}_b=0~$are the only the possible values which can be taken to satisfy the pink eqn. And from the definition of linear independence, it can be concluded that$~{\boldsymbol x_1\over\Vert\boldsymbol{x}_{1}\Vert},~\boldsymbol x_2~$are linearly independent.
