The inequality displayed in ($*$) is incorrect.
Consider $A=\mathbb{Z}$, $a=\frac{1}{2}$, $b=\frac{3}{2}$. An interval of length $\frac{1}{2}$ contains at most one integer, so for every $x\in\mathbb{R}$, $A\cap[x,x+a]$ is either empty, or has cardinality $1$. In particular, since there are $x$ for which the intersection is empty, then it follows that $\inf_{x\in\mathbb{R}}\mathrm{card}(A\cap [x,x+a]) = 0.$ On the other hand, an interval of length $\frac{3}{2}$ contains always at least one integer, and sometimes two; so $\inf_{x\in\mathbb{R}}\mathrm{card}(A\cap[x,x+b]) = 1.$ Therefore $\inf_{x\in\mathbb{R}}\mathrm{card}(A\cap [x,x+a]) = 0 \not\geq 1 = \inf_{x\in\mathbb{R}}\mathrm{card}(A\cap[x,x+b]).$
On the other hand, if you meant whether $\inf_{x\in\mathbb{R}}\mathrm{card}(A\cap[x,x+a])\leq\inf_{x\in\mathbb{R}}\mathrm{card}(A\cap [x,x+b]),$ then that inequality does hold.
Since $A$ is countable, the infimum on the left is either infinite, or a natural number.
If it is infinite, then $A\cap[x,x+a]$ is infinite for every $x$, and therefore $A\cap[ x,x+b]$ is infinite for every $b$, so both infima are infinite, hence equal.
Now suppose that the infimum on the left is finite, equal to $n$. For every $x$, $A\cap [x,x+a]\subseteq A\cap [x,x+b]$, so $\mathrm{card}(A\cap [x,x+a])\leq\mathrm{card}(A\cap [x,x+b])$. Since the infimum is less than or equal to every element of the set, we have $n\leq \mathrm{card}(A\cap [x,x+a])\leq\mathrm{card}(A\cap [x,x+b]),$ hence $n$ is a lower bound for the set $\{\mathrm{card}(A\cap[x,x+b])\mid b\in\mathbb{R}\},$ and hence we have $n\leq \inf_{x\in \mathbb{R}}\mathrm{card}(A\cap [x,x+b]).$ Thus, the desired inequality holds.