Is $| \lceil \frac{a}{2} \rceil - \lceil \frac{b}{2} \rceil |\geq \lfloor |\frac{a - b}{2}| \rfloor $? Let $a$ and $b$ be integers. Is it true that
$$
 \left | \left \lceil \frac{a}{2} \right \rceil -  \left \lceil \frac{b}{2} \right \rceil \right |\geq  \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor
$$
Where $\lceil \cdot \rceil$ is the ceiling function, $\lfloor \cdot \rfloor$ the floor function and $|\cdot|$ is the absolute function.
The inequality seems to be true when I check it programatically but I would like to get a proof (or disproof) for this inequality.

Assume without loss of generality that $a\ge b$. Then the inequality is
$$
\left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil
\ge \left\lfloor \frac {a-b}2 \right\rfloor
$$ 
If either $a$ or $b$ is an even integer, then we can pull the whole number $\frac a2$ or $\frac b2$ out of the floor function, and the inequality reduces to $\left\lceil \frac a2 \right\rceil \ge \left\lfloor \frac {a}2 \right\rfloor$ or $-\left\lceil \frac b2 \right\rceil \ge \left\lfloor -\frac {b}2 \right\rfloor$ (where the first is trivial and the second is actually an equality). 
Assume therefore that neither of $a$ and $b$ is an even integer. Let $2m<a<2(m+1)$ and $2n<b<2(n+1)$, for some $m,n\in \mathbb Z$. Then
$$
\left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil
= (m+1)-(n+1)
= m-n
$$
On the other hand
$$
m-n-1<\frac a2 - \frac b2 < m-n+1
$$
which means that 
$$
\left\lfloor \frac {a-b}2 \right\rfloor
\le m-n
= \left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil
$$
so we are done.
EDIT: I didn't notice you assumed $a$ and $b$ to be integers. Well, my answer works for all real numbers.
