Exercise 6.27 from Introduction to probability (Anderson,Seppalainen,Valko)

I am having trouble coming up with a solution to this problem. This is a recommended exercise for an upcoming midterm (Not for marks, just for practice).

$$\textbf{The question is:}$$

Suppose $$X_{1}$$ and $$X_{2}$$ are independent random variables with $$P(X_{1}=1)=P(X_{1}=-1)=\frac{1}{2}$$ and $$P(X_{2}=1)=1-P(X_{2}=-1)=p$$ with $$0. Let $$Y=X_{1}X_{2}$$. Show that $$X_{2}$$ and $$Y$$ are independent.

$$\textbf{My attempt at a solution}$$

Since $$X_{1},X_{2}$$ are independent, we have $$P(x_{1},x_{2})=P(x_{1})P(X_{2})$$. Therefore we have the following:

$$P_{X_{1},X_{2}}(-1,-1)=\frac{1-p}{2}$$

$$P_{X_{1},X_{2}}(-1,1)=\frac{p}{2}$$

$$P_{X_{1},X_{2}}(1,-1)=\frac{1-p}{2}$$

$$P_{X_{1},X_{2}}(1,1)=\frac{p}{2}$$

Which is in fact a pdf. Now, since $$Y=X_{1}X_{2}$$, we have $$(X_{1},X_{2}) \longrightarrow Y$$

$$(-1,-1) \longrightarrow Y=1$$

$$(-1,1) \longrightarrow Y=-1$$

$$(1,-1) \longrightarrow Y=-1$$

$$(1,1) \longrightarrow Y=1$$

Therefore, $$P_Y(1)=\frac{1}{2}$$ and $$P_Y(-1)=\frac{1}{2}$$.

$$\textbf{I am not entirely sure where to go from here (If any of this is even right at all)}$$

After finding pmf of $$Y$$, let $$x,y\in \{1,-1\}$$,
\begin{align} P(X_2 = x, Y=y) &= P(X_2=x, X_1=x_2y) \\ &= P(X_2=x)P(X_1=x_2y)\\ &= P(X_2=x) \cdot \frac12\\ &= P(X_2=x)P(Y=y) \end{align}