The Restriction of a Metric to a Subset Induces the Subspace Topology

Let $(X,d)$ be a metric space, let $A\subseteq X$, and let $\mathbb{R}$ be the set of real numbers. Let $d_A:A\times A\to\mathbb{R}$ be the restriction of $d$, that is, the function with $d_A(a,b)=d(a,b)$ for all $a,b\in A$. Equip $X$ with the collection $\mathcal{T}_d$ of all subsets open in $(X,d)$, which is a topology by Metric Open Sets Form a Topology, and let

$$\mathcal{T}_A=\{A\cap U: U\in\mathcal{T}_d\}$$

be the subspace topology on $A$. Then the following hold.

1. (Restriction is a metric) $d_A$ is a metric on $A$, so that $(A,d_A)$ is a metric space.

2. (Agreement of topologies) A subset $V\subseteq A$ is open in the metric space $(A,d_A)$ if and only if $V\in\mathcal{T}_A$. Consequently the collection of subsets of $A$ that are open in $(A,d_A)$ is exactly $\mathcal{T}_A$.