Say we want to show that the interior of a set $A$ is open.
If $x \in Int(A)$, then there exists an open ball $B_r(x) \subseteq A$.
Since $B_r(x)$ is open, $y \in B_r(x)$ also has an open ball $B_s(y) \subseteq B_r(x) \subseteq A$, so $y \in Int(A)$.
Now, somehow we have to show that the ball $B_r(x) \subseteq Int(A)$, and that would complete the proof.
All of the proofs I read say this is obvious, but I don't see how $B_r(x) \subseteq Int(A)$ immediately follows here.
Best Answer
$$\text{Int}(A)=\{x\in X\mid \exists r>0: B_r(x)\subset A\}.$$
Therefore $$\forall x\in \text{Int}(A), \exists r_x>0: B_r(x)\subset A,$$ and thus $$\text{Int}(A)=\bigcup_{x\in \text{Int}A}B_{r_x}(x).$$
We finally conclude that $\text{Int}(A)$ is open.