Could someone clarify the countability part in Munkres' proof that the ordered square is not path connected (Munkres Topology, Chapter 24, Example 6) 

The proof proceeds as follows:

Let $p = 0 \times 0$ and $q = 1 \times 1$. Suppose there is a path $f: [a,b] \to I\_o^2$ joining p and q. 

By the intermediate value theorem, $f([a,b])$ must contain every point $x \times y$ of $I\_o^2$. So, for each $x \in I$, define $U\_x = f$$^-$$^1(x \times (0,1))$. The sets $U\_x$ are non-empty, open in [a, b] by continuity of $f$, and disjoint.

Choose for each $x \in I$ a rational number $q\_x \in U\_x$. Thus since $U\_x$ are disjoint, the map $x \mapsto q\_x$ is an injective mapping of $I$ onto $\Bbb Q$ thus contradicting that the interval $I$ is uncountable. 

My question is how $x \mapsto q\_x$ can be injective if it's taking $x \in [0,1] \subset \Bbb R$ to $q\_x \in (0,1) \subset \Bbb Q$.