Can you expand induction proofs to the real numbers?

To add on to Rob Arthan's answer: the induction principle does not work if you put the quantifiers differently, that is: for all $x$ there is some $\epsilon > 0$ such that if $S_x$ holds, then also $S_y$ for all $y\in \mathbb{R}$ with $\lvert x-y\rvert < \epsilon$. The problem is that now the $\epsilon$ you get might get smaller and smaller when you try to go away from $x_0$ and you don't get very far. A counterexample could be the statement $$S_x = x \in (-1,1),$$ which is certainly true for $x = 0$ and around each $x\in (-1,1)$ you can find a small interval that is still completely contained in $(-1,1)$.

You have to be quite careful where you put the quantifiers in your principle, to make it valid. If you have $S_{x_0}$ and, for some $\varepsilon > 0$, you have that for all $x, y \in \Bbb{Q}$ (or $\Bbb{R}$) $S_x$ implies $S_y$ whenever $|x - y| < \varepsilon$, then you have that $S_x$ for all $x \in \Bbb{Q}$ (or $\Bbb{R}$). This is really just induction over the natural numbers for the proposition for $n \in \Bbb{N}$ that $S_x$ holds for all $x$ with $|x - x_0| < n\epsilon$.

The general replacement for induction in $\Bbb{R}$ is the principal of completeness: every non-empty set that is bounded above has a least upper bound. One application of completeness is to prove the Archimedean property: if $\epsilon > 0$, then for any $x$ there is $n \in \Bbb{N}$ such that $x < n\epsilon$ (the proof begins by assuming that $x$ is an upper bound for $\{n\epsilon \mid n \in \Bbb{N}\}$ and then derives a contradiction using completeness). The Archimedean property justifies the line of argument in the previous paragraph.