Worked solution: Gödel and Cohen: the continuum hypothesis is independent of ZFC (1963)

The condensation lemma: if $X$ is an elementary substructure of $L_\alpha$ (for $\alpha$ a sufficiently large limit ordinal), its Mostowski collapse is again exactly $L_\beta$ for some ordinal $\beta \le \alpha$. Combined with the downward Löwenheim–Skolem theorem, one can, for any $x \subseteq \omega$ with $x \in L$, find a countable elementary substructure $X \prec L_\gamma$ (for $\gamma$ large enough that $x \in L_\gamma$) with $x \in X$; collapsing $X$ gives some $L_\beta$ with $\beta < \omega_1$ (since $X$ is countable) and $x \in L_\beta$.

Hence every real of $L$ appears already in $L_{\omega_1}$, so $\mathcal{P}(\omega) \cap L \subseteq L_{\omega_1}$, and since $|L_{\omega_1}| = \aleph_1$, this gives $|\mathcal{P}(\omega) \cap L| \le \aleph_1$. Cantor's theorem already gives $\aleph_1 \le 2^{\aleph_0}$, so inside $L$, $2^{\aleph_0} = \aleph_1$: that is, $L \models \mathrm{CH}$ (the same argument, run at every cardinal $\aleph_\alpha$ instead of just $\aleph_0$, in fact gives the stronger generalized continuum hypothesis in $L$).

Since $L$ is built inside any model of $\mathrm{ZFC}$ using only the $\mathrm{ZFC}$ axioms themselves, this shows $\mathrm{Con}(\mathrm{ZFC}) \implies \mathrm{Con}(\mathrm{ZFC} + \mathrm{CH})$: $\mathrm{CH}$ can never be refuted from $\mathrm{ZFC}$ alone. This is only half of independence; the remaining three steps build a universe with strictly more reals than $L$ allows, to show $\neg\mathrm{CH}$ is equally consistent.