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.

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\begin{align}
& \color{#44f}{{1 \over x} + {1 \over 1 -x} > 0}
\sr{x\ \not\in\ \braces{0,1}}\implies
x^{2}\pars{1 - x}^{2}\
\overbrace{{1 \over x\pars{1 - x}}}
^{\ds{= 1/x + 1/\pars{1 - x}}} > 0
\\[5mm] & \implies x\pars{1 - x} >0 \implies
x\pars{x - 1} < 0 \implies
\bbx{\color{#44f}{0 < x < 1}}
\end{align}
