Like the first answer, this is really a comment that has outgrown the comment box. No new results here, but a different way of looking at continuous selectors that may be helpful (somewhat analogous to the relationship between choice functions and preference relations in decision theory).

Call a relation $T$ on a topological space $X$ an **open tournament on $X$** (thanks for the terminology!) if the following conditions hold:

(a) $T$ is total: for all $x,y \in X$, either $xTy$ or $yTx$;

(b) $T$ is antisymmetric: for all $x,y \in X$, if $xTy$ and $yTx$ then $x = y$;

(c) $T$ respects the topology on $X$: for all distinct $x,y \in X$ with $xTy$, there exist disjoint open neighbourhoods $U$ and $V$ about $x$ and $y$, respectively, such that for all $x' \in U$ and all $y' \in V$, $x'Ty'$.

Let's read $xTy$ and "$x$ trounces $y$" (yes, I am having some fun with this).

**Lemma:** The existence of a continuous selector for a Hausdorff space $X$ is equivalent to the existence of an open tournament on $X$.

*Proof.*
Given an open tournament $T$ on $X$, define $s: [X]^{2} \to X$ by
$$s(\{x,y\}) = x \iff xTy.$$
Conditions (a) and (b) ensure that $s$ is a well-defined selector function. Moreover, if $s(\{x,y\}) = x$ and $W$ is an open neighbourhood about $x$, then since $xTy$ we can find $U$ and $V$ as in condition (c), in which case the open sets $U \cap W$ and $V$ yield an open neighbourhood about $\{x,y\}$ contained in $s^{-1}(W)$, so $s$ is continuous. Note also that conditions (a) and (c) imply that $X$ is Hausdorff.

Conversely, suppose that $X$ is a Hausdorff space and $s$ is a continuous selector function. Define $T \subset X \times X$ by:
$$xTy \iff s(\{x,y\}) = x.$$
It is clear that $T$ is total and antisymmetric. Suppose that $x \neq y$ and $xTy$, so we have $s(\{x,y\}) = x$. Let $W$ and $W'$ be disjoint neighbourhoods of $x$ and $y$, respectively, and let $U$ and $V$ be the open neighbourhoods of $x$ and $y$ corresponding to $s^{-1}(W)$. Then $W \cap U$ and $W' \cap V$ are as required in condition (c). $\blacksquare$

Given an open tournament $T$ on $X$, for each $x \in X$ define
$$T_{x} := \{y \in X : x \neq y \textrm{ and } xTy\}$$
and
$$T^{x} := \{y \in X : x \neq y \textrm{ and } yTx\}.$$
Then it is easy to see that for all $x \in X$, $T_{x}$ and $T^{x}$ are open, and moreover
$$T_{x} \sqcup T^{x} = X - \{x\}.$$
This shows that a necessary condition for the existence of an open tournament on $X$ is that $X$ is disconnected whenever it is punctured. In particular, this provides another way of seeing that neither $\mathbb{R}^{2}$ nor $S^{1}$ admit continuous selectors. This condition is certainly not sufficient, however, since it is satisfied by the triod. In that case, something more subtle seems to be going on; loosely speaking, the problem seems to be that distinct punctures in a neighbourhood of the vertex create very different separations.

**Edit:** Let me try to make this last point a bit more precise. First some intuition. If $A$ and $B$ constitute a separation of a space $Y$, then $\chi_{A}: Y \to \{0,1\}$, the characteristic function of $A$, is both surjective (since $A \neq \emptyset$ and $A \neq Y$), and also continuous. Conversely, every continuous surjective characteristic function $\chi$ on $Y$ yields the separation $\chi^{-1}(1) \sqcup \chi^{-1}(0)$ of $Y$.

The product topology on the set $2^{Y}$ tells us that $\chi$ and $\chi'$ are "$F$-close" if they agree on the finite set $F \subset Y$ (these are the basic opens). This in turn provides a notion of closeness on the set
$$\mathcal{S}_{Y} := \{(A,B) : \textrm{$A$ and $B$ constitute a separation of $Y$}\}$$
of all (ordered) separations of $Y$: $(A,B)$ and $(A',B')$ are "$F$-close" if $A \cap F = A' \cap F$ (so also $B \cap F = B' \cap F$).

More generally, given a fixed Hausdorff space $X$, let
$$\mathcal{D} = \mathcal{D}_{X} := \{(A,B) : \textrm{$A$ and $B$ are open and disjoint}\},$$
and topologize this set using the notion of "$F$-closeness" as above: given a finite set $F \subset A \cup B$, say that $(A,B)$ and $(A',B')$ are "$F$-close" if $A \cap F = A' \cap F$ and also $B \cap F = B' \cap F$.

To connect this reasoning to the question at hand, observe that an open tournament $T$ on $X$ yields a function $f: X \to \mathcal{D}$ by setting $f_{T}(x) = (T_{x}, T^{x})$.

**Lemma:** $f_{T}$ is continuous.

*Proof.* Consider the basic open neighbourhood of $f_{T}(x) = (T_{x}, T^{x})$ corresponding to the finite set $F \subset T_{x} \cup T^{x} = X - \{x\}$. If $z \in T_{x} \cap F$, then in particular $xTz$ so there are open sets $U_{z}$ and $V_{z}$ about $x$ and $z$ respectively as in condition (c). Similarly, if $z \in T^{x} \cap F$ then $zTx$ so there are open sets $U_{z}$ and $V_{z}$ about $z$ and $x$ respectively as in condition (c). Set
$$W := \bigcap_{z \in T_{x} \cap F} U_{z} \cap \bigcap_{z \in T^{x} \cap F} V_{z}.$$
Then $W$ is an open neighbourhood of $x$ and for any $y \in W$ we have
$$T_{y} \cap F = T_{x} \cap F \textrm{ and } T^{y} \cap F = T^{x} \cap F,$$
which shows that $f$ is continuous. $\blacksquare$

Now we can see (from another perspective) why the triod admits no open tournament $T$: if it did and $v$ is the vertex point of the triod, then either $T_{z}$ consists of two of the prongs and $T^{z}$ consists of the third, or vice versa. However, in either case, $f_{T}$ cannot be continuous, since by choosing a point $w$ very close to $v$ (but distinct from $v$), we can always ensure that any two prongs lie in the same connected component, so we can always "ruin" whatever arbitrary choice was made at the vertex by $T$.

Here is a quick attempt at a converse. Let $f: X \to \mathcal{D}$ satisfy:

- $f$ is continuous;
- for each $x \in X$, $f(x) \in \mathcal{S}_{X - \{x\}}$;
- for all $x,y \in X$, $y \in (\pi_{0} \circ f)(x)$ iff $x \notin (\pi_{0} \circ f)(y)$.

Call such an $f$ (provisionally) "nice". One can check that when $T$ is an open tournament, $f_{T}$ is "nice". Define a binary relation $T_{f}$ on $X$ by:
$$xT_{f}y \iff y \in (\pi_{0} \circ f)(x).$$
Then I believe/hope that $T_{f}$ is an open tournament (though I still need to check this carefully).

A converse like this would be appealing because it provides a nice picture of when there exists a continuous selector. For example, $\mathbb{R}$ would have one due to the "nice" function
$$f(x) = (\{y : y < x\}, \{y : y > x\}).$$
Perhaps of more interest, the space $\mathbb{Q}^{2}$ would admit a continuous selector based on the existence of the following "nice" function: for each $p \in \mathbb{Q}^{2}$, let $L_{p}$ denote the subset of $\mathbb{Q}^{2}$ to the left of the line through $p$ of slope $\pi$, let $R_{p}$ denote the subset of $\mathbb{Q}^{2}$ to the right of the line through $p$ of slope $\pi$, and set $f(p) = (L_{p}, R_{p})$.

Follow-up: Is this equivalent to the linear ordering condition? – François G. Dorais♦ Sep 6 '11 at 0:50doeshave a continuous selector, as will any other countable metric space. – François G. Dorais♦ Sep 6 '11 at 4:14