Why isn't this a counter-example $Var(X_1 + X_2) = \sigma^2 + \sigma^2$ for independent $X_i$ s.t. $Var(X_i) = \sigma^2$?

Question

1 Hypothesis:

Let $X_1$ and $X_2$ be independent random variables on $\{-1, 1\}$ s.t.

$$ X_1(-1) = X_2(-1) = -1 $$

and

$$ X_1(1) = X_2(1) = 1. $$

Furthermore, let each event on $\{-1, 1\}$ be equiprobable.

Then the mean of $X_1$ and $X_2$ is $0$, and the variance on each is

$$ \sigma^2 = (-1)^2 0.5 + (1)^2 0.5 = 1 $$

2 Question:

Evidently, we have that $Var(X_1 + X_2) = \sigma^2 + \sigma^2 = 2$. But why isn't it in fact $4$ since

$$ (X_1 + X_2)(-1) = -1 + -1 = -2 $$

and

$$ (X_1 + X_2)(1) = 1 + 1 = 2 $$

so that

$$ Var(X_1 + X_2) = (-2)^2 \cdot 0.5 + (2)^2 \cdot 0.5 = 4? $$

3 Same question asked a different way

I agree that

$$ Var(kX)= k^2 \sigma^2 $$

So with that being the case, why don't we similarly have:

$$ Var(X_1 + X_2) = Var(2X_1) = Var(2X_2) = 2^2 \sigma^2 = 4 \sigma^2 = 4? $$

Answer 1

To be independent, you need $E(X_1X_2)=E(X_1)E(X_2)$. But $X_1X_2=1$ because both rely on the underlying $\{-1,1\}$ in the same way.
To be independent, they each need their own choice of $-1$ or $1$. Then you won't write $(X_1+X_2)(-1)$, but $X_1(-1)+X_2(-1)$ or $X_1(-1)+X_2(1)$ and so on.