Consider the integral $$I=\int_{-1}^{1}\frac{1}{x^2+1}\mathrm{d}x$$ Now, from the standard integral results we know, $$\int\frac{1}{x^2+1}\mathrm{d}x=\arctan(x)+c$$ So, $$\int_{-1}^{1}\frac{1}{x^2+1}\mathrm{d}x=\arctan(1)-\arctan(-1)=\frac{\pi}{4}-(-\frac{\pi}{4})=\frac{\pi}{2}$$ Now, if you are bored and just doing random things you might evaluate this integral in a roundabout way by the substitution $u=\dfrac{1}{x}$.
This gives the bound for the integrals as $u=\dfrac{1}{-1}=-1$ to $u=\dfrac{1}{1}=1$ and the differential becomes $\mathrm{d}{x}=\dfrac{-1}{u^2}\mathrm{d}{u}$. So, the integral becomes $$\int_{-1}^{1}\frac{1}{\frac{1}{u^2}+1}\frac{-1}{u^2}\mathrm{d}u = -\int_{-1}^{1}\frac{1}{u^2+1}\mathrm{d}u=-I$$.
Now, $$I=-I \implies 2I=0\implies \pi=0$$.
So, there is your false proof.
I think that the problem in the proof comes when we do the substitution $x=\dfrac{1}{u}$. But, I dunno what it is. Does it have to do something with the continuity of the substitution, or is it something else.
What you should actually get is that $$I=\int_0^1 \frac1{x^2+1}dx+\int_{-1}^0 \frac1{x^2+1}dx$$ Then the substitution will give $$I=-\int_\infty^1 \frac1{u^2+1}du-\int_{-1}^{-\infty} \frac1{u^2+1}du$$