Is T Always Equal to 2 in the Persistent Number Theorem?

[anemone]

Here is this week's POTW:

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Call a number $T$ persistent if the following holds:

Whenever $a,\,b,\,c,\,d$ are real numbers different from $0$ and $1$ such that

$a+b+c+d=T$ and $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}=T$

we also have

$\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}+\dfrac{1}{1-d}=T$

Prove that $T$ must be equal to $2$ if $T$ is persistent.

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[anemone]

Congratulations to Olinguito for his correct solution, which you can find below:

Spoiler

Let $T$ be a persistent number.

Choose real numbers $a,b\ne0,1$ such that
$$a+\frac1a+b+\frac1b\ =\ T.$$
We can always do this. For, observe that if $f(x)=x+\dfrac1x$, then $f(x)>2$ for all $x>1$ and $f(x)\to\infty$ as $x\to\infty$, meaning that given any $y>2$, there exists an $x>1$ such that $f(x)=y$. Also note that $f(x)\to-\infty$ as $x\to-\infty$.

Therefore, given the fixed number $T$, we can also choose an $a<0$ such that
$$a+\frac1a\ <\ T-2$$
and then there will be a $b>1$ such that
$$b+\frac1b\ =\ T-a-\frac1a\ >\ 2.$$

Thus we have
$$a+b+\frac1a+\frac1b\ =\ T\ =\ \dfrac1a+\dfrac1b+a+b.$$
Hence, as $T$ is persistent, we have
$$T\ =\ \frac1{1-a}+\frac1{1-b}+\frac1{1-\frac1a}+\frac1{1-\frac1b}$$
which simplifies to $\boxed{T=2}$.