Here is an example from Bhargav Bhatt's talk "Using DAG" at MSRI last week. Needless to say, any mistakes are mine.

**Theorem.** Let $X$ be a coherent (quasi-compact and quasi-separated) scheme, let $A$ be a ring complete with respect to an ideal $I\subseteq A$. Then
$$ X(A) \to \varprojlim_n X(A/I^{n+1}) $$
is bijective.

Before going into the proof, let us consider the case $X$ is affine. Then
$$
X(A) = {\rm Hom}(\Gamma(X, \mathcal{O}_X), A) = \varprojlim_n {\rm Hom}(\Gamma(X, \mathcal{O}_X), A/I^{n+1}) = \varprojlim_n X(A/I^{n+1}) . $$

The idea for the general (coherent) case is to replace $\Gamma(X, \mathcal{O}_X)$ with ${\rm Perf}(X)$, the category of perfect complexes on $X$.

**Slogan.** Affine schemes have "enough functions". Coherent schemes have "enough vector bundles (perfect complexes)".

The second idea may be due to Thomason.

More precisely, we have:

**Proposition.** Let $X$ and $Y$ be schemes.

(a) If $X$ is affine, then
$$ {\rm Hom}(Y, X) \to {\rm Hom}(\Gamma(X, \mathcal{O}_X), \Gamma(Y, \mathcal{O}_Y)) $$
is bijective.

(b) If $X$ is coherent, then
$$ {\rm Hom}(Y, X) \to {\rm Hom}({\rm Perf}(X), {\rm Perf}(Y)) $$
is an equivalence.

We must specify what (b) means (here is where DAG enters the picture). We consider ${\rm Perf}(X)$ as the symmetric monoidal $\infty$-category of perfect complexes on $X$ (complexes locally quasi-isomorphic to a bounded complex of locally free sheaves of finite rank). The ${\rm Hom}$ on the right means the $\infty$-groupoid (space) exact $\otimes$-functors. So in particular (b) implies that this space is discrete. The map in (b) sends $f$ to $f^*$, the pull-back functor.

*"Proof" of Theorem.* We repeat the proof of the affine case, replacing rings with categories of perfect complexes:
$$
X(A) = {\rm Hom}({\rm Perf}(X), {\rm Perf}(A)) = \varprojlim_n {\rm Hom}({\rm Perf}(X), {\rm Perf}(A/I^{n+1})) = \varprojlim_n X(A/I^{n+1}) . $$
Unlike in the affine case, the middle equality needs some justification, which I am not ready to give.

**End remarks.**

(1) I think Bhargav mentioned that an idea due to Gabber allows one to get rid of the assumption that $X$ is coherent in the Theorem.

(2) He also said that the above proof is the only one he is aware of.

(3) Reference for the above (thanks to the user *crystalline*):

Bhargav Bhatt *Algebraization and Tannaka duality* arxiv.org/abs/1404.7483.