First of all, despite claims by several people here, there are models of computation which support computation on topological spaces. The "countable vs. uncountable" distinction is a red herring, arising from being familiar only with notions of computation on discrete spaces. A good introductory text on the topic of computable analysis is Klaus Weihrauch's Comuptable analysis – an introduction.
Your idea is very closely related to the principle of open induction, which states: if an open subset $S \subseteq [0,1]$ satisfies $(\forall y \in [0,1].\, y < x \Rightarrow y \in S) \Rightarrow x \in S$ then $S = [0,1]$. In words: if an open $S$ has the property "if all numbers smaller than $x$ are in $S$ then so is $x$" then $S$ is the whole interval.
It is easy enough to speculate a functional version of open induction, which we could call open recursion (caveat – I am speaking off the top of my head here): given a map $$h : \Pi (x : [0,1]) \, (f : [0,x) \to \mathbb{R}) .\, \Sigma (\epsilon > 0) (g : [0, x + \epsilon) \to \mathbb{R}) . \, g{\restriction}_{[0,x)} = f$$ there is a unique map $\mathcal{O} h : [0,1] \to \mathbb{R}$ such that
$
\mathcal{O} h (t) = h \, t \, ((\mathcal{O} h){\restriction}_{[0,t)})
$
for all $t : [0,1]$, where $f{\restriction}_{[0,t)}$ is the restriction of $f$ to the interval $[0,t)$.
The principle of open induction holds in classical mathematics (exercise), from which it follows that open recursion is valid as well (another exercise). That is, the notation $\mathcal{O} \, h$ is well-defined for any $h$.
Now, the interesting question is whether we can compute with open recursion. In An induction principle over real numbers explained how to compute with open induction. I am fairty sure that her method can be adapted to open recursion, but it would take more time to write up the details than I can invest in answering this question. I would be very interested to hear, however, if my hunch is correct.