Hensel's lemma can be stated as
Let $A$ be a local ring complete by the maximal ideal $m$, and define $Q=A/m$. Let $f\in A[x]$ be a monic polynomial and define $\bar{f}\in Q[x]$ as the polynomial where the coefficients are reduced mod $m$. If we have monic polynomials $G,H\in Q[x]$ with $(G,H)=1$ such that $\bar{f}=GH$, then we have monic polynomials $g,h\in A[x]$ such that $f=gh$, $\bar{g}=G$ and $\bar{h}=H$.
I'm having trouble with completing the proof. So far I've proved by induction that for $i\geq 1$ we have $G_i, H_i\in A[x]$ such that $f=G_iH_i\ mod\ m^i[x]$ where $\overline{G_i}=G$ and $\overline{H_i}=H$, with $\deg(G_i)=\deg(G)$ and $\deg(H_i)=\deg(H)$.
I then showed that coefficient wise $G_i$ and $H_i$ are Cauchy, and thus converge to coefficients $a_0,\ldots,a_{deg(G_i)-1}$ and $b_0,\ldots, b_{deg(H_i)-1}$ for $G_i$ og $H_i$. Then if we let \begin{align*} g(x)=x^{deg(G_i)}+a_{deg(G_i)-1}x^{deg(G_i)-1}+\ldots+a_1x+a_0 \\ h(x)=x^{deg(H_i)}+b_{deg(H_i)-1}x^{deg(H_i)-1}+\ldots+b_1x+b_0 \end{align*} we have $\overline{g}=G$ og $\overline{h}=H$.
Now I need to show that $f=gh$, but I'm kinda stuck. It's not clear to me, why (or how) we have $f=gh$.
Can you use what you know about the various $G_j$, $H_j$ to show that $f = gh \mod m^i[x]$ for each $i$? (Hint: show that the $G_j$ etc are unique. That is, if $f = G_j H_j = G'_j H'_j \mod m^j[x]$, then $G_j = G'_j, H_j = H'_j$.)
Once you have this, then you know that $f - gh \in m^i[x]$ for each $i$, and hence $f - gh \in \bigcap_i m^i[x]$, i.e. the $x^k$-coefficient of $f - gh$ is in $\bigcap_i m^i$ for each $k$.
It remains to show that the $m$-adic filtration is separated on $A$ (which is a fancy way of saying $\bigcap_i m^i = 0$).
Recall that your local ring $(A,m)$ is complete, which Matsumura uses to mean that the natural map $\varphi: A\to \underset{\leftarrow n}{\lim} A/m^n$ is an isomorphism. (We'll only need it to be injective, actually.) Take any $x\in \bigcap_i m^i$: then the image of $x$ in each $A/m^n$ is $0$, and so $\varphi(x) = 0$, which implies that $x=0$ as $\varphi$ is injective.