I want to show that the ring $\mathbb{Q}_p$ is a field for any prime $p$, so I want to show that every nonzero element has an inverse. I thought of the following argument, but I can't seem to locate where I am using the fact that $p$ is prime. So instead of $p$, let me use $6$ throughout although I never use anything really special about $6$. Exercise 12 on p.210 of Stein's Algebraic Number theory book here claims that $\mathbb{Q}_6\simeq \mathbb{Q}_2 \times \mathbb{Q}_3$, so $\mathbb{Q}_6$ isn't a field.
Let $\alpha:=\{x_n\}_{n>0}$ be a nonzero element of $\mathbb{Q}_6$ (thought of as the completion of $\mathbb{Q}$ with respect to the 6-adic absolute value), i.e the $x_n \in \mathbb{Q}$ and they form Cauchy sequence with respect to the 6-adic valuation $|-|_6$. Since $\alpha$ is nonzero, we have $|\alpha| \neq 0$, i.e $\lim\limits_{n \to \infty}|x_n|_6 \neq 0$. Hence there exists $\epsilon>0$ and $N$ such that $|x_n|_6>\epsilon$ for all $n>N$. In particular $x_n \neq 0$ for $n>N$, and hence $x_n^{-1} \in \mathbb{Q}$ for $n>N$.
We now construct the inverse $\beta$ of $\alpha$. Let $\beta=\{y_n\}$ be the sequence given by $y_n=x_n^{-1}$ for $n>N$, and for $0<n<N$, let $y_n$ be anything. I have to show the $\{y_n\}$ are a Cauchy sequence in order to conclude $\beta \in\mathbb{Q}_6$. $$x_n^{-1} - x_m^{-1} = \frac{x_m-x_n}{x_nx_m}$$ Since $\{x_n\}$ is Cauchy, given $\epsilon_1>0$, there exists $N_1$ such that $|x_m - x_n|_6 < \epsilon_1\epsilon^2$ for $m,n>N_1$. Also, $|x_mx_n|_6 >\epsilon^2$ from the first paragraph for $m, n >N$. Hence $|x_n^{-1} - x_m^{-1}|_6<\epsilon_1$, for $m,n>\max\{N_1, N\}$. So the $y_n$ form a Cauchy sequence, and hence $\beta \in \mathbb{Q}_6$. Finally, $\alpha\beta=1 \in \mathbb{Q}_6$ since the nth-term of $\alpha\beta$ is 1 for $n>N$.
So my question is, what is wrong with this argument? or if you can tell me a correct proof of why $\mathbb{Q}_p$ is a field for $p$ prime? Or give an element of $\mathbb{Q}_6$ with no multiplicative inverse?
Your proof is fine for primes, but doesn't work for composites. The problem is that the $m$-adic absolute value is not multiplicative if $m$ is composite. For instance, $|2|_6=|3|_6=1$ but $|2\cdot 3|_6=1/6$. This means that in your argument, you cannot conclude from $|x_n|_6,|x_m|_6>\epsilon$ that $|x_nx_m|_6>\epsilon^2$.
To get a counterexample, you can inductively choose convergent sequences of integers $(x_n)$ and $(y_n)$ such that $x_n$ is divisible by higher and higher powers of $2$ but not divisible by $3$, and $y_n$ is divisible by higher and higher powers of $3$ but not divisible by $2$. Then the limits $x$ and $y$ will be nonzero, and in fact have valuation $1$, since none of the $x_n$s or $y_n$s are ever divisible by $6$). But $xy=0$ because $x_ny_n$ is getting more and more divisible by $6$, so $x$ and $y$ cannot be invertible.