Yes, it is true.
$$
\left | \left \lceil \frac{a}{2} \right \rceil - \left \lceil \frac{b}{2} \right \rceil \right |\geq \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor
\tag1$$
In the following, $m,n$ are integers.
Case 1 : If $a=2m,b=2n$, then both sides of $(1)$ equal $|m-n|$.
Case 2 : If $a=2m,b=2n+1$, then
$$(1)\iff |m-n-1|\ge \left\lfloor\left |m-n-\frac 12\right|\right\rfloor\tag2$$
If $m-n-\frac 12\ge 0$, then $m-n-1\ge 0$, so$$(2)\iff m-n-1\ge m-n-1$$which is true.
If $m-n-\frac 12\lt 0$, then $m-n-1\lt 0$, so$$(2)\iff -m+n+1\ge -m+n$$which is true.
Case 3 : If $a=2m+1, b=2n$, then
$$(1)\iff |m-n+1|\ge \left\lfloor\left|m-n+\frac 12\right|\right\rfloor\tag3$$
If $m-n+\frac 12\ge 0$, then $m-n+1\ge 0$, so$$(3)\iff m-n+1\ge m-n$$which is true.
If $m-n+\frac 12\lt 0$, then $m-n+1\lt 0$, so$$(3)\iff -m+n-1\ge -m+n-1$$which is true.
Case 4 : If $a=2m+1,b=2n+1$, then both sides of $(1)$ equal $|m-n|$.
Hint:
First notice, that when $x$ is an integer, the inequality does not hold.
Let's write $x=n+\alpha$, where $n$ is an integer and $0 < \alpha < 1$, then we can rewrite the inequality as $(n+\alpha)^2 < n(n+1)$. Now the problem is reduced to solving the following inequality:
$$\alpha^2 + 2n \alpha - n < 0$$
Can you take it from here?
Best Answer
HINT:
As $\displaystyle \binom nk>0$ for $0\le k\le n$ where $n>0,k$ are integers
Check for $k$ such that $$\frac{\binom n{k+1}}{\binom nk}=\frac{n-k}{k+1}>=<1$$