This is probably not as impressive as other answers, but since I am quite fond of real induction, I could not resist.
Real induction can be stated as follows:
Let $a < b$ be real numbers. Let a subset $S \subseteq [a,b]$ be inductive, i.e.:
(RI1) $a\in S$.
(RI2) If $a\le x<b$, then $x\in S$ $\implies$ $[x,y]\subseteq S$ for some $y > x$.
(RI3) If $a < x \le b$ and $[a,x)\subset S$, then $x \in S$.
Then $S=[a,b]$.
This formulation is taken from this Pete L. Clark's answer where also his text on this topic is linked.1
If we look at the set $S'=\{x\in[a,b]; [a,x]\subseteq S\}$, then the above conditions simply say that $S'$ is non-empty clopen subset of $[a,b]$. So in this way, we can view this as a consequence of connectedness.
The linked paper contains several theorems shown using real induction. Usually the proof goes by choosing a suitable set $S$ and showing that this set fulfills the conditions (RI1), (RI2), (RI3). At least for some of them it seems that there is not much difference between difficulty of the proof that $S$ is a non-empty clopen subset and the proof that $S$ fulfills these conditions. (Of course, it is more elegant to use real induction if we want to illustrate this method as unifying principle for proofs of several basic theorems. And also in this way similarity to the usual mathematical induction is highlighted. But for the purposes of this question, this approach might be a way how to view proofs by real induction as a source of applications of connectedness in analysis.)
1 Pete L. Clark: The Instructor's Guide to Real Induction, https://arxiv.org/abs/1208.0973, http://math.uga.edu/~pete/realinduction.pdf