$X$, $Y$ independent identically distributed. Are there counterexamples to symmetry of $X-Y$? That $X-Y$ should be symmetrically distributed for iid $X,Y$ is obvious simply by interchanging the roles of $X$ and $Y$ -- informally we might argue

Let $Z=X-Y$ have distribution $F$. The roles of which observation was called $X$ and which $Y$ is arbitrary; therefore $-Z=Y-X$ must have the same distribution. If $-Z$ and $Z$ have the same distribution then the distribution is symmetric (about $0$).

However I have a vague recollection of having encountered an odd counterexample at some point (I wonder if it might perhaps have been in the Counterexamples book by Romano and Siegel).
Is there some subtlety in the above outlined argument (symmetry in the roles of $X$ and $Y$ implies symmetry of the distribution function) that goes astray in some edge case, or is the implied more formal version of the argument solid? (indicating that I am simply misremembering the notional existence of an exception)
I can't see any obvious way to break it but "I don't see how to do it" doesn't mean much as an argument. I suppose that it may be I have misrecalled some detail; perhaps the exception actually resulted because there wasn't independence in the original formulation (in which case I believe I could find an exception myself).  [Edit: Indeed, I have done so now]
I expect the answer is "no, you're misremembering, obviously it's symmetric" but this nags at me now and then and I worry my notions are flawed in some way.
 A: Corrected after @Glen_b pointed out a glaring error. Sloppy proof, but should work.
I think we can prove this using characteristic functions. 
Let X, Y be iid. Let Z = Y-X Then,
$\phi_{X-Y}(t) = E[e^{it(X-Y)}] = \phi_X(t)\phi_{-Y}(t)$.
Similar to the CDF, the characteristic function of X uniquely characterizes the distribution of X, and it exists for any real-valued random variable. This implies that $\phi_X(t) \equiv \phi_Y(t)$. 
This, along with properties of the characteristic function under linear transformation implies that
$\phi_{-X}(t) =\phi_{X}(-t)  = \phi_{Y}(-t)=\phi_{-Y}(t)  $.
In turn, this implies that
$\phi_X(t)\phi_{-Y}(t) = \phi_X(t)\phi_{-X}(t) = \phi_Y(t)\phi_{-X}(t)  =\phi_{Y-X}(t) $,
so that $\phi_{X-Y}(t)=\phi_{Y-X}(t)$ and $Z \sim -Z$.
A: Just to clear up the source of my own confusion, I managed to coax just enough (about 4 lines!) out of Google books to resolve the origin of my doubt. It was from Romano and Siegel* and what they actually have there is:

4.34 Identically distributed random variables such that their difference does not have a symmetric distribution
If $X$ and $Y$ are independent and identically distributed, then $X-Y$ has a symmetric distribution about zero. The independence assumption cannot be dropped in general. (However, if $X$ and $Y$ are exchangeable, then $X-Y$ does have a symmetric distribution.)

A simple counterexample I came up with is $X\sim U[0,3)$ and $Y=(X-1) \text{ mod } 3$, so $Y$ is also uniform on $[0,3)$ and for which $X-Y$ takes the value $1$ with probability $\frac23$ and $-2$ with probability $\frac13$. This isn't the one I thought of just after I posted the question, but once you have one, further counterexamples are easy to think up and this one is simpler to explain. (Edit: in the end I managed to see their counterexample for the dependent case; it's fine - a simple bivariate example on $\{-1,0,1\}^2$ - but mine's simpler to express so I'll leave it there.)
Note here that $(X,Y)$ doesn't have the same distribution as $(Y,X)$ -- so we don't have the exchangeability that R&S mention, which is why the asymmetry is possible. Note also that the informal argument in my question - "interchange the roles of $X$ and $Y$" - quite directly relied on exchangeability and we can see from that outline why that weaker condition should be sufficient to get that $X-Y$ and $Y-X$ have the same distribution.
* Romano, J.P. and Siegel, A.F. (1986),
Counterexamples in Probability And Statistics, (Wadsworth and Brooks/Cole Statistics/Probability Series)
(my question was based on a recollection from reading a little of it in late 1986... so it's no surprise I was a little fuzzy on the details)
