Sum of independent symmetric random variables is symmetric This is exercise 3.2.5 from Probability and Random Processes by Grimmett and Stirzaker:
Let $X_r$, $1 \leq r \leq n$, be independent random variables which are discrete and symmetric about $0$, that is, $X_r$ and $-X_r$ have the same distributions. Show that for all $x$, $P(S_n \geq x) = P(S_n \leq -x)$ where $S_n = \sum_{r = 1}^n X_r$.
Is the following proof correct?
$\textbf{Proof:}$
Let $x \in \mathbb{R}$ be arbitrary. Then,
\begin{eqnarray}
P(S_n \geq x) & = & P\left(\sum_{r = 1}^n X_r \geq x\right) \\
& = & P\left(\underset{x_1 + \cdots + x_n \geq x}{\bigcup_{x_1, \ldots, x_n}} \{X_1 = x_1\} \cap \cdots \cap \{X_n = x_n\} \right) \\
& = & \underset{x_1 + \cdots + x_n \geq x}{\sum_{x_1, \ldots, x_n}} P\left(\{X_1 = x_1\} \cap \cdots \cap \{X_n = x_n\}\right) \\
& = & \underset{x_1 + \cdots + x_n \geq x}{\sum_{x_1, \ldots, x_n}} P\left(\{X_1 = x_1\}) \cdots P(\{X_n = x_n\}\right) \textrm{(from independence of variables)}\\
& = & \underset{x_1 + \cdots + x_n \leq -x}{\sum_{x_1, \ldots, x_n}} P\left(\{X_1 = -x_1\}) \cdots P(\{X_n = -x_n\}\right) \textrm{(from symmetry of variables)} \\
& = & \underset{x_1 + \cdots + x_n \leq -x}{\sum_{x_1, \ldots, x_n}} P\left(\{X_1 = -x_1\} \cap \cdots \cap \{X_n = -x_n\}\right) \textrm{(from independence of variables)}\\
& = & P\left(\underset{x_1 + \cdots + x_n \leq -x}{\bigcup_{x_1, \ldots, x_n}} \{X_1 = -x_1\} \cap \cdots \cap \{X_n = -x_n\} \right) \\
& = & P\left(\sum_{r = 1}^n X_r \leq -x\right) \\
& = & P(S_n \leq -x)
\end{eqnarray}
as required. $\square$
Is there a quicker/better way of proving the result?
 A: Everything seems correct. 
But yes, there is both quicker and better way, based on the fact (which is easy to prove) that a random variable is symmetric iff its characteristic function is real-valued. Given this, the symmetry of $S_n$ immediately follows from
$$
\varphi_{S_n} (t) = \prod_{k=1}^n \varphi_{X_k}(t).
$$
It is better because it allows to prove the required property for any random variables, not only for discrete.
A: For $x$ in the support of $X_1+\cdots+X_n$, let $S$ be the set of all $n$-tuples $(x_1,\ldots,x_n)$ in $\operatorname{support}(X_1) \times\cdots\times \operatorname{support}(X_n)$ for which $x_1+\cdots+x_n=x$, and let $T$ be the set of all $n$-tuples $(x_1,\ldots,x_n)$ in the former support for which $x_1+\cdots+x_n=-x$. Then
\begin{align}
& \Pr(X_1+\cdots + X_n = x) \\[10pt]
= {} & \sum_{(x_1,\ldots,x_n)\in S} \Pr(X_1 = x_1\ \&\ \cdots \ \&\ X_n=x_n) \\[10pt]
= {} & \sum_{(x_1,\ldots,x_n)\in S} \Pr(X_1 = -x_1\ \&\ \cdots \ \&\ X_n=-x_n) \\[10pt]
= & \sum_{(x_1,\ldots,x_n)\in T} \Pr(X_1 = x_1\ \&\ \cdots \ \&\ X_n=x_n) \tag 1 \\[10pt]
= & \Pr(X_1+\cdots+X_n= -x).
\end{align}
To justify the equality in $(1)$, you have to write an argument relying on the symmetry of each of the separate distributions.  And the reason you can think of them separately is independence.
