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.