Problem: Suppose that $p > 2$. I want to show that $x^2 \equiv 2($mod $p$) has a solution iff 2 has a square root in $\mathbb{Q}_p$.
My attempt: Suppose that $x^2 \equiv 2($mod $p$) has a solution, say $x_1$, then take $x_2 = x_1 + \delta_1 p$ so that $x_2^2 \equiv 2($mod $p^2$). Similarly, take $x_3 = x_2 + \delta_2p^2$ so that $x_3^2 \equiv 2 ($mod $p^2$). Repeat this process, we obtain a sequence $(x_n)$ such that $x_n^2 \equiv 2 ($mod $p^n$). So $\lim\limits_{n \to \infty} |x_n^2-2|_2 \leq \lim\limits_{n \to \infty} p^{-n} = 0$. Thus $x_n^2 \to 2$ and hence $x_n \to \sqrt{2}$. This shows that 2 has a square root in $\mathbb{Q}_p$.
Is this correct? I've only been introduced to the concept of $\mathbb{Q}_p$ in the context of metric completion. Is there any hint or suggestion for the backward direction?
Your approach works, but you need to justify that $x^2 \equiv 2 \pmod{p^n}$ has a solution if we only assume that $x^2 \equiv 2 \pmod{p}$ has a solution. Take a look at Hensel's lemma for this.
For the backwards direction, note that $1=|2|_p=|\sqrt{2}|_p^2$ implies $|\sqrt{2}|_p=1$, so $\sqrt{2} \in \Bbb Z_p$. Now reduce mod $p$ and use that $\Bbb Z_p/p\Bbb Z_p \cong \Bbb Z/p\Bbb Z$.