Infimum of two sets $\newcommand{\card}{\operatorname{card}}$
Let $A$ be a nonempty countable set of real numbers, and $0< a\leq b$. Is the following true:
$$ \tag{*} \inf_{x\in \mathbb R} \card(A\cap [x,x+a]) \geq  \inf_{x\in \mathbb R} \card(A\cap [x,x+b]) $$
where $\card$ means number of elements in the set.
As far as I know, since $a\leq b$ then 
$$A\cap [x,x+a] \; \subseteq  \; A\cap [x,x+b] $$
so
$$ \card(A\cap [x,x+a]) \; \leq\;  \card(A\cap [x,x+b]) $$ for all $x\in \mathbb R$. How I can complete the proof (if the result is correct)?
(I know that if we have two sets $B\subseteq C$ then $\inf B \geq\inf C$. But how I can use this in terms of cardinality of sets?)
 A: 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.
A: $\newcommand{\card}{\operatorname{card}}$The correct is
$$ \inf_{x \in \mathbb{R}}  \card(A\cap [x,x+a]) \le \inf_{x \in \mathbb{R}}\card(A\cap [x,x+b]).$$
By definition of $\inf$ there exists $x_n \in \mathbb{R}$ such that 
$$\card(A\cap [x_n,x_n+b]) \le \inf_{x \in \mathbb{R}}\card(A\cap [x,x+b]) + 1/n.$$
Hence, 
$$ \inf_{x \in \mathbb{R}}  \card(A\cap [x,x+a]) \le \card(A\cap [x_n,x_n+a]) \le \card(A \cap [x_n,x_n+b]) $$
$$\le \inf_{x \in \mathbb{R}}\card(A\cap [x,x+b]) +1/n.$$
Do $n \rightarrow \infty.$
