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 $\sqrt{2}\in\mathbb{Q_7}$?

This is how I see it and, perhaps, it will help you: we can easily solve the polynomial equation
$$p(x)=x^2-2=0\pmod 7\;\;(\text{ i.e., in the ring (field)} \;\;\;\Bbb F_7:=\Bbb Z/7\Bbb Z)$$
and we know that there's a solution $\,w:=\sqrt 2=3\in \Bbb F_{7}\,$
Since the roots $\,w\,$ is simple (i.e., $\,p'(w)\neq 0\,$) , Hensel's Lemma gives us lifts for the root, meaning: for any $\,k\ge 2\,$ , there exists an integers $\,w_k\,$ s.t.
$$(i)\;\;\;\;p(w_k)=0\pmod {7^k}\;\;\wedge\;\;w_k=w_{k-1}\pmod{p^{k-1}}$$
Now, if you know the inverse limit definition of the $\,p$-adic integers then we're done as the above shows the existence of $\,\sqrt 2\in\Bbb Q_p\,$ , otherwise you can go by some other definition (say, infinite sequences and etc.) and you get the same result.
