Theorem. The set of interior points of any set $A$, written $\operatorname{int}(A)$, is an open set.
So I know for proving any set to be open, we take a arbitrary element in that set and want to choose a radius at which open sphere of that radius lies completely in that set.
Let $p\in \operatorname{int}(A)$, then by definition of int($A$), there exist an open sphere of $p$ with radius $r$, $S_{r}(p)\subset A$.
But we want a radius $r_1$ around $p$, such that $S_{r_{1}}(p)\subset \operatorname{int}(A)$. Now my doubt starts.
Ok, we know that $\operatorname{int}(A)$ is always a subset of $A$, so we want to reduce the radius $r$ to $r_1$ , but for that , procedure the book chosen is beyond my understanding even we try to imagine it by making diagrams.
Please check the attached picture and if someone help me in understanding this, I shall be very thankful to him/her. 
Best Answer
The proof from the picture is wrong, because $z \in S_r(x)$ does not imply that $z \in \operatorname{Int} A$. It is also needlessly complicated. A correct proof is as follows:
Let $x \in \operatorname{Int} A$ so there is some $r > 0$ satisfying $S_r(x) \subseteq A$. It suffices to show that $S_r(x) \subseteq \operatorname{Int} A$. So take any $y \in S_r(x)$ and put $r_1 = r - d(x, y)$. Then $S_{r_1}(y) \subseteq S_r(x)$ since for $z \in S_{r_1}(y)$ we have that
$$d(z, x) \leqslant d(z, y) + d(y, x) < r_1 + d(y, x) = r.$$
It follows that $S_{r_1}(y) \subseteq S_r(x) \subseteq A$, so $y \in \operatorname{Int} A$ as desired. $\square$
Side note: denoting the sets $S_r(x)$ and calling them spheres is awkward, since the standard definition involves open balls which are usually denoted $B_r(x)$.