If $XY$ is degenerate and $X, Y$ are independent, then $X, Y$ are degenerate. 
Let $X$ and $Y$ be independent random variables such that
  $$P(XY=c)=1,$$
  for some $c\neq 0$. Why are $X$ and $Y$ necessarily also degenerate? That is, I want to prove that there exists $a,b\in\mathbb{R}$ such that
  $$P(X=a)=1,\quad\text{and}\quad P(X=b)=1.$$

It seems hard because there is no way in which we can tell what $a$ and $b$ will be. Any numbers such that $ab=c$ could work. So I have to prove they exist without finding them explicitly. How to do that?
 A: Suppose, to make things a little simpler, that $X$, $Y$ are non-negative. Let $Z=XY$ and $c >0$, and $F_X(x) = P(X \le x)$ (cumulative distribution function).
Then, for any given $z \ge 0$, and for any $x\ge0 $:
$$P(X \le x \cap Y \le z/x) \le P(Z \le z) \le  P(X \le x \cup Y \le z/x) $$
Or (here we use independence) :
$$ F_X(x) F_Y(z/x) \le F_Z(z) \le F_X(x) + F_Y(z/x) - F_X(x) F_Y(z/x)$$ 
Now, because $Z$ is totally degenerate, $F_Z(z)$ has a unit step at $z=c$:
$$ F_Z(z) = \cases{0, \; z<c\\
1 , \; z \ge c}$$
If $F_X(x)$ has a unit step at $x=a$, then it's straighforward that $F_Y(y)$ has a step at $y=b=c/a$.
That is, if $X$ is totally degenerate, so is $Y$ - and viceversa.
Let's assume that's not the case, ($X$ is not totally degenerate). Then there must exist some $a>0$ such that $F_X(a)=\alpha$ with $0<\alpha<1$. Let $b=c/a$
Then, for any $z < c$ we have $F_Z(z)=0 \ge \alpha \, F_Y(z/a)$ . This implies that $F_Y(y)=0$ for all $y<b$.
Further, if $z \ge c$, we have $F_Z(z)=1 \le F_Y(z/a)(1-\alpha)+\alpha$. This implies 
$F_Y(y)=1$ for all $y\ge b$. So, $Y$ is totally degenerate, and hence $X$ must also be, which contradicts the assumption.
