There are several relatively straightforward existence proofs in analysis that make use of $c < 2^c.$
1. "Most" Lebesgue measure zero sets are not Borel sets, because there are $2^c$ many Lebesgue measure zero sets (consider all subsets of a measure zero Cantor set) and there are only $c$ many Borel sets.
2. "Most" Riemann integrable functions are not Borel measurable, because the characteristic function of any subset of a measure zero Cantor set is Riemann integrable and there are only $c$ many Borel measurable functions.
3. "Most" complete Borel measures on $\mathbb R$ are not $\sigma$-finite. In fact, there are $2^c$ many complete Borel measures on $\mathbb R$ and only $c$ many $\sigma$-finite Borel measures (complete or not complete) on ${\mathbb R}.$ To see the first claim, let $B$ be a Borel set of cardinality $c$ (e.g. $B$ could be a Cantor set or the interval $[0,1]).$ For each $A \subseteq B,$ define ${\mu}_A(E) = \infty$ if $A \cap E \neq \emptyset$ and ${\mu}_A(E) = 0$ if $A \cap E = \emptyset.$ To see the second claim, note that every finite Borel measure on $\mathbb R$ is the Lebesgue-Stieltjes measure of some monotone function, and there are only $c$ many monotone functions (several ways to prove this). Now observe that every $\sigma$-finite Borel measure on $\mathbb R$ can be associated with a sequence of finite Borel measures on ${\mathbb R}.$ (Recall that there are only $c$ many sequences whose terms all come from a given set of cardinality $c.)$
4. "Most" convex subsets of ${\mathbb R}^2$ are not Borel sets, since removing any subset of the boundary of the unit disk results in a convex set and there are only $c$ many Borel sets. Note how badly this fails for ${\mathbb R}.$
5. "Most" functions $f:{\mathbb R} \rightarrow {\mathbb R}$ that are symmetrically continuous at each point (i.e. for each $x \in \mathbb R$ we have $\lim_\limits{h\rightarrow 0}\ [f(x+h)-f(x-h)]=0)$ are not continuous, or even Borel measurable. Miroslav Chlebík proved in this 1991 Proc. AMS paper that there are $2^c$ symmetrically continuous functions, and there are only $c$ many continuous functions (indeed, only $c$ many Borel measurable functions).
6. "Most" subsets of the boundary of the unit disk are not a divergence set for any power series with complex coefficients and radius of convergence $1,$ since there are $2^c$ many subsets of the boundary of the unit disk and only $c$ many power series with complex coefficients. For more details about the possible divergence sets of a power series with complex coefficients, see this answer. Note how different this is for power series with real coefficients, in which there are only $2^2 = 4$ possible subsets of the boundary of an interval (there are only $4$ subsets of a $2$-element set) and it is not difficult to see that any of these subsets can be a divergence set.