Proving $\sqrt{2}\in\mathbb{Q_7}$? Why does Hensel's lemma imply that $\sqrt{2}\in\mathbb{Q_7}$?
I understand Hensel's lemma, namely: 

Let $f(x)$ be a polynomial with integer coefficients, and let $m$, $k$ be positive integers such that $m \leq k$. If $r$ is an integer such that $f(r) \equiv 0 \pmod{p^k}$ and $f'(r) \not\equiv 0 \pmod{p}$ then there exists an integer $s$ such that $f(s) \equiv 0 \pmod{p^{k+m}}$ and $r \equiv s \pmod{p^{k}}$.

But I don't see how this has anything to do with $\sqrt{2}\in\mathbb{Q_7}$?
I know a $7$-adic number $\alpha$ is a $7$-adically Cauchy sequence $a_n$ of rational numbers. We write $\mathbb{Q}_7$ for the set of $7$-adic numbers. 
A sequence $a_n$ of rational numbers is $p$-adically Cauchy if $|a_{m}-a_n|_p \to 0$ as $n \to \infty$.
How do we show $\sqrt{2}\in\mathbb{Q_7}$?

Hensel's lemma allows you, for simple roots of polynomial equations, to get from an integer solution of the equation modulo $p^k$ do a solution modulo $p^{k+m}$, which is called a "lift" of the solution because reduction modulo $p^k$ will project back to the original solution. I don't know why the condition $m\leq k$ is given in the statement of the lemma in Wikipedia, since by simple iteration one can get this for $m$ as large as one likes. Then repeating this indefinitely, one gets the $p$-adic Cauchy sequence you want to have.
Concretely, if you start with the solution $n=3$ modulo $p^1=7$ to the equation $n^2=2$ defining $\sqrt2$, then you can find a unique solution $n'$ of that equation modulo $7^2=49$ that also "projects back" to $n$ modulo $7$, that is $n'\equiv 3\pmod 7$, by writing $n'=3+7k$ and then solving for $k$ in $(n')^2\equiv2\pmod{49}$. This becomes $3^2+2\times3\times7k+49k^2\equiv2\pmod{49}$ or $42k=-7\pmod{49}$, which after division by $7$ gives $6k\equiv-1\pmod7$, which has the solution $k=1$, unique modulo$~7$, and $n'\equiv3+7\times1=10$ satsifies both $(n')^2\equiv2\pmod{49}$ and $n'\equiv 3\pmod 7$. 
Playing the game again posing $n''=n'+49l$ we get $(n'')^2=10^2+2\times10\times49l+10^2\times49^2l^2$, which we can compute modulo $49^2=7^4$ dropping the final term, and given that $10^2-2=98$ is divisible by $49$ (we ensured this above) one can divide the congruence $(n'')^2\equiv 2\pmod{7^4}$ by $49$ to get $20l\equiv-98/49=-2\pmod{7^2}$, and since $20$ is invertible modulo $49$, with inverse $125$, this has the solution $l\equiv =125\times-2=-250\equiv44\pmod{49}$, unique modulo $49$, giving $n''=2166$. This can be continued indefinitely (the explicit computations become harder to perform, but the reason they succeed remains the same), giving solutions modulo ever higher powers of $7$. Then $2,10,108,2166,\ldots$ gives a $7$-adic Cauchy sequence converging $7$-adically to a square root of $2$ (that is congruent to $3$ modulo $7$). Note that I've added the reduction $108$ of $2166$ modulo $7^3$, although its presence does not make the Cauchy sequence any better (nor of course does the presence of any particular term in the sequence) just to mark the fact that we've jumped over the step of solving modulo $7^3$; one could of course just retain in the Cauchy sequence the values that were actually computed.
