Placing Ts on the $x$-axis 
A "T" consists of two perpendicular intervals $\{c\}\times[0,a]$ and $[b,d]\times \{a\}$ (with $b<c<d$) on the plane. We say that the T is placed on point $c$. Is it possible to place non-intersecting T's on all real numbers on the $[0,1]$ interval of the $x$-axis?

I believe the answer should be "no". Suppose it were possible. We can assume that each T has equal left-halfwidth and right-halfwidth (i.e. half of the horizontal line of the T.) For each real number $x\in[0,1]$, let $h_x$ denote the height of its T and $w_x$ denote its halfwidth. Then two real numbers $x,y\in[0,1]$ have intersecting T's if $h_y< h_x$ and $|x-y|\le w_y$, and vice versa. So every time we have $h_x>h_y$, we must have $|x-y|>w_y$. How can we get a contradiction?

This may not actually address your question, but it may help you think about it in a different way:
For any set of finitely many $x_i \in [0,1]$, we know you can construct a set of open intervals $I_i$ such that $x_i \in I_i$ and $x_i \notin I_j \;\forall j < i$. This is (should be) fairly easy to understand why; a really easy way to construct all of these intervals is if we put all our $x_i$'s in ascending order, assign the first $x$ the interval $(-\infty,\infty)$, and each subsequent $x_i$ the interval $(x_{i-1}, \infty)$. It is clear to see that we can show this for the case for any finite sequence of $x$ values.
To show how we manage this for an infinite number of values, I am going to put the sequence together like this: $\frac{1}{2}, \frac{1}{4}, \frac{3}{4}, \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}...$
We can then see that we can assign each number an interval like this: $(-\infty,\infty),(-\infty,\frac{1}{2}),(\frac{1}{2},\infty),(-\infty,\frac{1}{4}),(\frac{1}{4},\frac{3}{4}),(\frac{3}{4},\infty)...$
I'm not quite sure what this sort of sequence is called, but if we tile them this way we, in addition showing we can do this for all numbers, we get a really cool fractal structure.
This has a lout of similarity to what the problem describes, and you ay be able to do the same thing with the T's. (except for the problem that the T's have closed intervals...)
