Yes, $x$ is a real number. But it's not any real number. By the time $x$ is identified, the number $A$ has already got a value. Suppose, for instance, that $a = 7$, and $A = 2$. Then $x$ is not just any real number ... it's a real number between $7-2$ and $7+2$, i.e.,
5 < x < 9.
The claim is that if you have such a number -- say $x = 5.1$ -- then there's a number $P$, less than $A$, with the property that $x$ lies between $a - P$ and $a + P$. In this example, the number $P = 1.95$ would work, because $5.1$ lies between $7 - 1.95 = 5.05$ and $7 + 1.95 = 8.95$.
In short, if $x$ lies in an open interval around $a$ of width $4$, then it also lies in an open interval of some width a little less than $4$.
Note that if we were talking about closed intervals, we'd have
5 \le x \le 9
and $x = 5$ is a possible value. But then you cannot reduce the size of the interval. So the open-ness is essential to what you're trying to show.