$\operatorname{int}(A)$ is open for any set $A$ of points in a metric space 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. 
 A: 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 $B_r(x)$.
A: A cleaner approach is to show $S_r(x)$ is an open set for all $x \in X$ and $r>0$, as a lemma of sorts.
This is in fact what they're really doing from line 5 onwards, anyway.
To prove the lemma follow the book's proof. (So we have fixed but arbitrary $x \in X$ and $r>0$ from now on).
To see that $S_r(x)$ is open we need to show that every arbitrary point $y \in S_r(x)$ is an interior point of $S_r(x)$: so having such an arbitrary $y$ we note that by definition of $S_r(x)$ we have $d(x,y) < r$. Then define the "wiggle room" $r' = r-d(x,y) >0$. The claim is then that this $r'$ "witnesses" that $y$ is an interior point of $S_r(x)$ which means
$$S_{r'}(y) \subseteq S_r(x)\tag{1}$$
To prove $(1)$ we show the inclusion by picking any $z \in S_{r'}(y)$. Again we know $d(z,y) < r'$. We need that $z \in S_r(x)$ so we want to show $d(x,z) < r$ by definition.
Now we use the triangle inequality (via the only other point that we know anything about, our old friend $y$):
$$d(x,z) \le d(x,y) + d(y,z) < d(x,y) + r' = r$$
as $r'$ was defined as $r-d(x,y)$ from the start. So we have what we wanted: $d(x,z) < r$ and $z \in S_r(x)$ and $(1)$ holds.
As all points were arbitrary, $S_r(x)$ is open.

Now the proof about $\operatorname{int}(A)$ being open: it's the same procedure: to show a set is open we pick any $x \in \operatorname{int}(A)$ and show it is an interior point of $\operatorname{int}(A)$. Knowing $x \in \operatorname{int}(A)$ tells us there is some $r>0$ such that $S_r(x) \subseteq A$. The claim is that in fact $$S_r(x) \subseteq \operatorname{int}(A)\tag{2}$$ and we are done: the same $r$ witnesses that $x$ is an interior point of $\operatorname{int}(A)$ as well. To see $(2)$ (an inclusion) we pick $y \in S_r(x)$ and show $y \in \operatorname{int}(A)$. We use the lemma above that $S_r(x)$ is open so $y$ is an interior point of it and so $r'>0$ exists so that $S_{r'}(y) \subseteq S_r(x)$. So $$y \in S_{r'}(y) \subseteq S_r(x) \subseteq A$$ and so $y \in \operatorname{int}(A)$ and $(2)$ has been shown. QED.
