Why is every smooth quartic in $\Bbb{P}^3$ a K3 surface?

First note that we have the short exact sequence $$0 \rightarrow \mathscr{O}_{\mathbb{P}^3}(-4) \rightarrow \mathscr{O}_{\mathbb{P}^3} \rightarrow \mathscr{O}_S \rightarrow 0.$$ You get what you want then by considering the long exact sequence of cohomology groups plus the fact that $H^1(\mathbb{P}^3,\mathscr{O}_{\mathbb{P}^3}) = H^2(\mathbb{P}^3,\mathscr{O}_{\mathbb{P}^3}(-4)) = 0$.

Edit:

Since you were asking for some references, I would recommend these lecture notes or their newer (and different) version. In general I can only recommend the two lecture notes full notes (older notes) and very recent full notes by Gathmann. I think everything is well motivated, explained and contains detailed examples etc. He will also explain all the things you need to understand my answer, i.e. the long exact cohomology sequence associated to a short exact sequence and the computation of the cohomology of the twisted sheaves on projective spaces.

Use Hodge theory $$H^1(S,\mathcal{O})=H^{0,1}(S)\subset H^1(S,\mathbb C),$$

together with Lefschetz hyperplane theorem $$H^1(S,\mathbb Z)\cong H^1(\mathbb P^3,\mathbb Z)=0.$$