TITLE: Why is Markov's Principle an axiom?
QUESTION [6 upvotes]: Markov's Principle: Let $ x \in \mathbb{R}$. Then the following holds:
\begin{align*}
\neg (x = 0) \Longrightarrow \vert x \vert >0.
\end{align*}
In constructive mathematics (no law of excluded middle) this is viewed as an axiom, which follows from the Limited Principle of Omniscience.
But isn't there also a constructive proof for the MP?
For example, in my opinion the following holds:
\begin{align*}
\neg (x = 0 ) \Longrightarrow \frac{1}{x} x = 1 \Longrightarrow \vert \frac{1}{x} x \vert = \vert  1 \vert > 0 \Longrightarrow \vert \frac{1}{x} \vert > 0 \wedge \vert x \vert > 0.
\end{align*}
Moreover, for $\neg (x=0)$ let $n \in \mathbb{N}$. Then we either have (by approximate splitting)
\begin{align*}
\vert x \vert > \frac{1}{n}
\end{align*}
or 
\begin{align*}
\vert x \vert < \frac{1}{2n}.
\end{align*}
In the first case, MP holds. In the latter case, letting $ n \longrightarrow \infty$, yields $\vert x \vert = 0$, hence $x = 0$ which is a contradiction.
Obviously my "proof" has to be wrong at some point, but I don't see where.

REPLY [3 votes]: The problem with your "proof" is the use of $\frac{1}{x}$. What notion of field are you using?
If you are using a discrete field, then it is justified. But since the reals aren't a discrete field, you probably aren't.
If you are using a Heyting field or weaker, then you can only talk about the inverse if you can establish that $x$ is apart from $0$, which is most likely equivalent to $|x|>0$. Likely another way of reformulating this variant of MP is $\neg(x=0)\implies x\# 0$ using $\#$ for apartness.