Interior of a set property

Question

$\DeclareMathOperator{\int}{int}$Let $A,B$ be subspaces of the metric space $X$. If $A \subseteq B$ then $\int A \subseteq \int B$.

Proof:

Let $x \in\int A$. Since $x$ is interior to the subspace $A$, there exists a radius $r>0$ so that $B(x,r) \subseteq A$. Because $A \subseteq B$, it must be the case that $B(x,r) \subseteq B$ and so $x$ is interior to $B$, in other words, $\int A \subseteq \int B$.

I would very much love feedback on this proof.

Answer 1

Good job!

Anyway observe that, $\text{Int}A \subseteq A \subseteq B$. So $\text{Int}A$ is an open set contained in $B$. But $\text{Int}B$ is the largest open set contained in $B$, so we must have $\text{Int}A \subseteq \text{Int}B$