Placing Ts on the $x$-axis

Here is an analytic approach to the problem. See my other answer for a different approach.

Define a few functions: $r:[0,1]\to\mathbb R$ assigning $z$ the value $d$ of the $T$ on $z$; similarly, $l:[0,1]\to\mathbb R:z\mapsto b$. Also, define $h:[0,1]\to\mathbb R$ by assigning $x$ the value $a$ of the $T$ at $x$.

Note that no $x$ satisfies $h(x)=0$. Define $U=\{x\in[0,1]:h(x)<0\}$.

Theorem. $U$ is dense in $[0,1]$.
Proof. Let $(x,y)$ be an open interval in which all $T$'s are upright--$h(z)>0$ for all $z\in(x,y)$. Set $\delta=\frac{y-x}3$. Pick any point $z_0\in(x,x+\delta)$. Define inductively $z_{n+1}$ by allowing it to be any point satisfying $z_n<z_{n+1}<\min\{c_n,z_n+2^{-(n+1)}\delta\}$, where $c_n$ is the minimum of the various $r_m,m\le n$. Then $\langle z_n\rangle$ is an increasing sequence bounded above by $x+2\delta$. It converges to $z\in(x,y)$.

Observe, $z\le r(z_n)$ for any $n$, so in order to avoid intersections $h(z)<h(z_n)$. Moreover, $z>l(z)>z_n$ for all $n$. Taking limits, $l(z)=z$, a contradiction. $\square$

Similarly, the set $V=\{x\in[0,1]:h(x)>0\}$ is also dense in $[0,1]$. We will now construct two sequences $\langle x_n\in U\rangle$ and $\langle y_n\in V\rangle$ in a fashion similar to the construction in the proof above. Pick $x_0<2^{-2}$ in $U$. Define $a_0=r(x_0)$, and choose $y_0\in V$ for which $x_0<y_0<\min\{a_0,x_0+2^{-3}\}$. Define $a_1=\min\{a_0,r(y_0)\}$. Given $x_n,y_n,a_{2n+1}$. Pick $x_{n+1}\in U$ such that $$y_n<x_{n+1}<\min\{a_{2n+1},y_n+2^{-(2n+3)}\}.$$ Let $a_{2n+2}=\min\{a_{2n+1},r(x_{n+1})\}$. Choose $y_{n+1}\in V$ such that $$x_{n+1}<y_{n+1}<\min\{a_{2n+2},x_n+2^{-(2n+4)}\}.$$ Let $a_{2n+3}=\min\{a_{2n+2},r(y_{n+1})\}$. Then, both sequences are bounded and increasing, and they have a shared limit $z$, which lies in $[0,1]$.

Since $z\le r(x_n)$ for every $n$, we can argue as in the proof of the theorem to get either $z\in V$ or $l(z)=z$. In the same way, either $z\in U$ or $l(z)=z$. Since $U\cap V=\emptyset$, it must be true that $l(z)=z$. This is a contradiction.