The proof goes as follows... "Given $x,y \in \mathbb{R}^n-A,$ Choose a point $z \in [x,y]$ other than $x$ and $y$, where $[x,y]=\{a \in \mathbb{R}^n :a=(1-t)x+ty,0\leq t\leq1\}$, Now choose $w \in \mathbb{R}^n-[x,y]$ such that $[z,w] \cap[x,y]=\{z\} $, ..."
To select such a $w$ I make a following claim and Proof.
KINDLY CHECK THAT THE CLAIM AND PROOF GIVEN BY ME BELOW IS CORRECT.
let $(x,y)=\{a \in \mathbb{R}^n :a=(1-t)x+ty, t \in \mathbb{R} \}$
Choose a point in $\mathbb{R}^n-(x,y)$ now, I claim that this point will do the job of $w$ required, It is easy too see by the geometry that my claim is correct...
But If I want to prove it rigorously, I need to prove the claim
Let $w \in \mathbb{R}^n-(x,y),z \in [x,y]$ then $[z,w] \cap [x,y]=\{z\}$
Proof:- Suppose not, then there exists $u \neq z$ such that $u \in [x,y]\cap[z,w]$
Since $u \in [x,y]$ and $u \in [z,w]$ there exists $r,s \in \mathbb{R}$ such that $u=(1-r)x+ry$ and $u=(1-s)z+sw$
from this I get $$sw=(1-r)x+ry-(1-s)z$$ $$\implies sw=u-(1-s)z$$ $$\implies w=\frac{u}{s}-\frac{1-s}{s}z$$Let $\frac{1}{s}=m$
$$\implies w=mu+(1-m)z$$
Which means that $w$ belongs to $(u,z)$ which is equal to $(x,y)$ and hence a c0ntradiction... Is this proof Legit?
Here's a different argument which is perhaps more visual.
Since $A$ is countable, you can enumerate the points $A=\{(a_{1,k},a_{2,k},...,a_{n,k})\}_{k\in\Bbb{N}}$
Now pick two points $x,y\in\Bbb{R}^{n}$ such that $x=(x_{1},...,x_{n})$ and $y=(y_{1},...,y_{n})$. Then atleast one $x_{i}$ and one $y_{j}$ must be not of the form $a_{i,k}$ or $a_{j,m}$ for any $m,k\in\Bbb{N}$ .
Suppose first that $i=j$ and that $x_{i}$ and $y_{i}$ are not members of the sequence $(a_{i,k})_{k\in\Bbb{N}}$ and suppose that $i\neq 1$
Then the line joining $(x_{1},...,x_{i},...,x_{n})$ and $(y_{i},y_{2},...,x_{i},...,y_{n})$ is in $\Bbb{R}^{n}-A$ as in this line, the i-th coordinate is always $x_{i}$ and hence is always in $\Bbb{R}^{n}-A$
Then you join the points $(y_{i},y_{2},...,x_{i},...,y_{n})$ and $(y_{i},...,y_{i},...,y_{n})$ by means of another line which will be in $\Bbb{R}^{n}-A$ as the first coordinate $y_{i}$ is still different from $a_{i,k}$ for each $k$.
Then you join the points $(y_{i},y_{2},...,y_{i},...,y_{n})$ and $(y_{1},...,y_{i},...,y_{n})=y$ and this is possible as again the $i$-th coordinate is different from the $a_{i,k}$'s.
Now suppose that $i$ and $j$ are different.
WLOG suppose that $x_{1}$ and $y_{2}$ are the coordinates which are different from $a_{1,k}$ and $a_{2,m}$ for each $k,m$.
Then you first go from $(x_{1},x_{2},...,x_{n})$ to $(x_{1},y_{2},...,y_{n})$ along the line which will be contained in $\Bbb{R^{n}}-A$ as $x_{1}$ is a coordinate different from $a_{1,k}$ Then you join $(x_{1},y_{2},...,y_{n})$ and $(y_{1},y_{2},...,y_{n})$ by means of another line which which will again be contained in $\Bbb{R}^{n}-A$ as $y_{2}$ remains constant along this line.
Argue by symmetry that this covers all cases.
Then the line joining $(x_{1},...,x_{i},...,x_{n})$ and $(x_{1},...,y_{i},...,x_{n})$ is in $\Bbb{R}^{n}-A$
Then you join the points $(x_{1},...,y_{i},...,x_{n})$ and $(y_{1},...,y_{i},...,y_{n})$ by means of another line which will be in $\Bbb{R}^{n}-A$ as $y_{i}$ is still different from $a_{i,k}$ for each $k$.