Looking for the flaw in the claim which this question asks about, one candidate is that the derivative of a function in $\mathbb{Q}_2$ is not necessarily itself in $\mathbb{Q}_2$. Is it?
$\mathbb{Q}_2$ is the 2-adic numbers.
Specifically, if $f$ is a function $f:\mathbb{Q}_2\to\mathbb{Q}_2$
Then is $\displaystyle\frac{df(x)}{dx}$ in $\mathbb{Q}_2$ for all $x\in\mathbb{Q}_2$ for all such $f$ ?
I guess I'm asking whether $\displaystyle\frac{df(x)}{dx}$ also satisfies $\displaystyle\frac{df(x)}{dx}:\mathbb{Q}_2\to\mathbb{Q}_2$
How about the definition: Let $f \colon \mathbb Q_2 \to \mathbb Q_2$ be a function. Let $a \in \mathbb Q_2$ be a point. We say $f$ is differentiable at $a$, and its derivative is $b$ iff $$ \lim_{x \to a}\frac{f(x)-f(a)}{x-a} = b $$ where the limit is done with the metric of $\mathbb Q_2$ and the quotient and difference are done with the operations of $\mathbb Q_2$.
If that is your definition, then (as in the real case) not every function is differentiable, but if a function is differentiable at some point, then the derivative indeed belongs to $\mathbb Q_2$.
left to you:
(1) Is $f(x) = x^2$ differentiable and $f'(x) = 2x$ ??
(2) Investigate differentiability of $\frac{1}{x}$.
(3) After you understand these, make a definition of "topological field" and try it there.