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For any point A and B in $\mathbb R^n-\{a\}$, I have to make a path connecting A and B.

Due to the $\{a\}$, I have to find 2 paths, to assure $\{a\}$ doesn't get in the way along one of the paths.

I thought straight line connecting A and B, but can't figure out about another path.

In my opinion, a ball containing A and B would be useful to imagine but...I can't rewrite that in the form of mathematics! (of course I can write the ball but don't know about writing path in mathematical form)

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    If you have two points $A$, $B$, then connect them by defining two straight lines to a point $z$ which is not $a$. The concatenation of this two paths gives the desired result. – 2017-02-04

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Let $x,y\in \Bbb{R}^n-\{a\}$. You're right that the straight line path works, if $a$ is not contained in the path joining $x$ and $y$.

What if $a$ is contained in this line? Choose another point $z$ (almost any point will work...) and connect $x$ to $z$, then $z$ to $y$ by straight lines. This gives you a path from $x$ to $y$.

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    There are in fact uncountably many such poygonal paths of two lines, such that each two of them only intersect at $x$ and $y$, but never in the middle. So if we omit countably many points, these paths can only intersect countable many of them, so there are infinitely many paths left that will work in the remainder of the space. – 2017-02-04
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In polar coordinates around {a}, a linear interpolation of the parameters works. For example, in 3D: $A = (r_a, \theta_a, \phi_a)$ and $B = (r_b, \theta_b, \phi_b)$. The path is: $$\left\{(1-q)(r_a, \theta_a, \phi_a)+q(r_b, \theta_b, \phi_b),\;\; 0\leq q\leq 1\right\}$$ This works because $r$ is always positive.