Hensel's lemma

Construct $\alpha$ as a limit of successive approximations $a = a_0, a_1, a_2, \dots \in \mathbb{Z}_p$ with $a_{n+1} = a_n - f(a_n)/f'(a_n)$, showing by induction that $v_p(f(a_n)) \ge n+1$ and $v_p(f'(a_n)) = v_p(f'(a)) = 0$ for all $n$ (the derivative stays a unit since $a_{n+1} \equiv a_n \pmod p$ at every step, so $f'(a_n) \equiv f'(a) \not\equiv 0 \pmod p$ throughout). Then $v_p(a_{n+1}-a_n) = v_p(f(a_n)) - v_p(f'(a_n)) \ge n+1$, so $(a_n)$ is Cauchy in the $p$-adic metric; by completeness of $\mathbb{Z}_p$ it converges to some $\alpha \equiv a \pmod p$, and continuity of $f$ forces $f(\alpha) = \lim f(a_n) = 0$. Uniqueness: if $\beta \ne \alpha$ were another root with $\beta \equiv a \pmod p$, the mean value / Taylor expansion $f(\beta) - f(\alpha) = (\beta-\alpha)(f'(\alpha) + p(\cdots))$ with $f'(\alpha)$ a unit would force $\beta = \alpha$.