Solving $\frac{1}{x} + \frac{1}{1-x} > 0$. Where's my error? 
For which $x$ is the following true?
$$\frac{1}{x} + \frac{1}{1-x} > 0$$

I’ve worked through the question a couple ways and got the correct answer, but the initial approach has a flaw and I’m not sure where in which step it is.
I got the correct answer by adding the two numbers to get $\frac{1}{x-x^2} > 0$ and going from there.
However my first attempt went like:
$$\begin{align}\frac{1}{x} &> - \frac{1}{1-x} \tag1\\[4pt]
\Rightarrow \qquad \frac{1-x}{x} &> -1 \tag2\\[4pt]
\Rightarrow \qquad 1-x &> -x \tag3 \\[4pt]
\Rightarrow \qquad x-1 &< x \tag4 
\end{align}$$
For which the answer is all $x$. However, the correct answer, which is clear from the first approach is $0 < x < 1$.
I’m new to working with inequalities and I figure there is just a simple property of them I’m being ignorant about which means one of my rearrangements isn’t true.
All help appreciated. Thanks.

"I figure there is just a simple property of them I’m being ignorant about which means one of my rearrangements isn’t true": that's it.
Let us take a few examples :

*

*$\color{blue}5<\color{green}8$ and $\color{blue}5\times2<\color{green}8\times2$

*$\color{blue}{-5}<\color{green}{\frac83}$ and $\color{blue}{-5}\times2<\color{green}{\frac83}\times2$

*$\color{blue}{-8}<\color{green}{-5.3}$ and $\color{blue}{-8}\times2<\color{green}{-5.3}\times2$
But

*

*$\color{blue}5<\color{green}8$ and $\color{blue}5\times{(-2)}\color{red}>\color{green}8\times(-2)$

*$\color{blue}{-5}<\color{green}{8}$ and $\color{blue}{-5}\times(-2)\color{red}>\color{green}{8}\times(-2)$

*$\color{blue}{-8}<\color{green}{-5}$ and $\color{blue}{-8}\times(-2)\color{red}>\color{green}{-5}\times(-2)$
