Solution:

a.
Suppose $T$ is persistent. Observe that for any $u$ satisfying $|u| \geq 2$, the equation $x + 1/x = u$ has real solutions (because it can be written in the form $x^2 - u x + 1 = 0$, which has discriminant $u^2 - 4$). Choose $u$ large enough so that both $u$ and $v = T - u$ have absolute value greater than $2$. Then there exist real $x, y$, not equal to $0$ or $1$, such that $x + 1/x = u$ and $y + 1/y = v$. Let $a = x$, $b = 1/x$, $c = y$, $d = 1/y$. Then $a + b + c + d = 1/a + 1/b + 1/c + 1/d = u + v = T$, while
$$\begin{aligned} T &= \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} + \frac{1}{1-d} \\ &= \frac{1}{1-x} + \frac{1}{1-1/x} + \frac{1}{1-y} + \frac{1}{1-1/y} \\ &= \frac{1}{1-x} + \frac{x}{x-1} + \frac{1}{1-y} + \frac{y}{y-1} \\ &= \frac{1-x}{1-x} + \frac{1-y}{1-y} \\ &= 2, \end{aligned}$$
and we are finished.

b.
Now we will show that $2$ is persistent.
Suppose $a + b + c + d = 2$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = 2$. Consider the monic polynomial $f(x)$ with roots $a, b, c, d$. Let $f(x)$ have expansion
$$f(x) = x^4 - e_1 x^3 + e_2 x^2 - e_3 x + e_4$$
By Vieta's formulas, we have
$$e_1 = a + b + c + d = 2$$
and
$$\frac{e_3}{e_4} = \frac{b c d + c d a + d a b + a b c}{a b c d} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = 2.$$
Thus, $f(x)$ has the form $x^4 - 2 x^3 + r x^2 - 2 s x + s$.
Next, we consider the polynomial $g(x) = f(1-x)$, which is the monic polynomial with roots $1-a, 1-b, 1-c, 1-d$. We have
$$\begin{aligned} g(x) &= (1-x)^4 - 2(1-x)^3 + r(1-x)^2 - 2s(1-x) + s \\ &= x^4 - 2 x^3 + r x^2 + (2 - 2r + 2s)x - (1 - r + s). \end{aligned}$$
By Vieta's formulas again, we have
$$\begin{aligned} \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} + \frac{1}{1-d} &= \frac{-(2 - 2r + 2s)}{-(1 - r + s)} \\ &= 2. \end{aligned}$$
Therefore, $2$ is persistent.