Let $X$ and $Y$ be i.i.d. random variables; we want to calculate the conditional expectation with respect to the $\sigma$-algebra generated by $X+Y$:
$$E [X \mid \sigma(X+Y)]$$
Now, generally for random variables $X, Y \in L^1$, if $$E[X1_A(X)1_B(Y)] = E[Y1_A(Y)1_B(X)]\ \quad (A, B \in \mathcal{B}(\mathbb{R}))$$
then$$E[X1_C(X+Y)] = E[Y1_C(X+Y)]\ \quad (C \in \mathcal{B}(\mathbb{R}))$$
So here is my solution so far: the above holds for i.i.d. random variables $X, Y$, so $$E[X \mid \sigma(X+Y)] = E[Y \mid \sigma(X+Y)]$$, and then we have $$E[X \mid \sigma(X + Y) ] = \frac{1}{2} E[X + Y \mid \sigma(X + Y) ] = \frac{X+Y}{2}$$
I feel like I am missing something here...
The third line in your proof is proved rigorously as follows: $$E[X1_C(X+Y)]=\int x1_{\{(x,y): x+y \in C \}} dF_{X,Y}(x,y).$$ Applying the transformation $(x,y)\to (y,x)$ and noting that $F_{X,Y}=F_{Y,X}$ we see that $$\int x1_{\{(x,y): x+y \in C \}} dF_{X,Y}(x,y) =\int y1_{\{(x,y): x+y \in C \}} dF_{X,Y}(x,y)$$. Hence $$E[Y1_C(X+Y)]=E[X1_C(X+Y)]$$.