This is an expansion on the comment I made to the effect that the symmetric square of $\mathbb{Q}^2$ has a continuous selector, which is surprising since the symmetric square of $\mathbb{R}^2$ does not have a continuous selector.

In fact, a much more general statement is true:

> The symmetric square of a countable metric space always has a continuous selector.

Let $X$ be a countable metric space. I will show that $X^2$ minus the diagonal can be split into two open sets $U$ and $V$ such that $(x,y) \in U$ iff $(y,x) \in V$. This is sufficient since $$\hat{s}(x,y) = \begin{cases} x & \text{when $(x,y) \in U$} \\\\ y & \text{otherwise} \end{cases}$$ is a continuous function $\hat{s}:X^2\to X$ such that $\hat{s}(x,y) = \hat{s}(y,x)$ for all $x,y \in X$. It follows that $\hat{s}$ is the lifting to $X^2$ of a continuous selector $s$ from the symmetric square of $X$ to $X$.

Let $d$ be a metric on $X^2$ such that $(x,y) \mapsto (y,x)$ is an isometry (e.g., let $$d((x_1,y_1),(x_2,y_2)) = d_0(x_1,x_2) + d_0(y_1,y_2)$$ where $d_0$ is a metric on $X$).
Since $X$ is countable, the set $D$ of all possible values of $d$ is also countable, which means that the set $E = (0,\infty) - D$ of non-values of $d$ contains arbitrarily small positive numbers.

Fix an enumeration $(x_0,y_0),(x_1,y_1),\ldots$  of $X^2$ minus the diagonal. We will define $U$ and $V$ by stages.

To start things off, let $\varepsilon_0 \in E$ be sufficiently small that the open ball $B_{\varepsilon_0}(x_0,y_0)$ does not intersect the symmetric ball $B_{\varepsilon_0}(y_0,x_0)$. (In particular, neither ball intersects the diagonal of $X^2$.) Put all points of $B_{\varepsilon_0}(x_0,y_0)$ in $U$ and all points of $B_{\varepsilon_0}(y_0,x_0)$ in $V$.

Next, we consider the point $(x_1,y_1)$. If $(x_1,y_1)$ was already put in $U$ or $V$, then skip to the next stage. Otherwise, let $\varepsilon_1 \in E$ be sufficiently small that the open ball $B_{\varepsilon_1}(x_1,y_1)$ does not intersect symmetric ball $B_{\varepsilon_1}(y_1,x_1)$, nor does either ball contain any points that were already put in $U$ or $V$. This is always possible since $\varepsilon_0$ is not a possible value of $d$, which means that $$\min\{d((x_0,y_0),(x_1,y_1)), d((y_0,x_0),(x_1,y_1))\} > \varepsilon_0.$$ So making sure that $$\varepsilon_1 < \min\{d((x_0,y_0),(x_1,y_1)),d((y_0,x_0),(x_1,y_1))\} - \varepsilon_0$$ will do for the second requrement. Finally, put all points of $B_{\varepsilon_1}(x_1,y_1)$ in $U$ and all points of $B_{\varepsilon_1}(y_1,x_1)$ in $V$.

Continue in the same manner for all the remaining points $(x_2,y_2),(x_3,y_3),\ldots$&nbsp;
Since all points of $X^2$ minus the diagonal will eventually be considered, in the end we will have a partition of $X^2$ minus the diagonal into two disjoint open sets $U$ and $V$ such that $(x,y) \in U$ iff $(y,x) \in V$.