# Restrict a metric, gives same topology as subspace topology from larger space $X$

Let $A\subseteq X$. If $d$ is a metric for the topology of $X$, show that $d\restriction_{A\times A}$ is a metric for the subspace topology on $A$.

I've shown that $d'=d\restriction_{A\times A}$ is a metric on $A$.

I am letting $\tau_B$ denote the subspace topology on $A$ that is induced by the metric topology on $X$ and $\tau_{B'}$ denote the topology on $A$ induced by the metric $d'$. My goal is to show $\tau_B=\tau_{B'}$.

Let $B=\{A \cap B_\varepsilon^d(x) \mid x \in X, \varepsilon>0\}$, a basis for $\tau_B$ and $B'=\{B_\varepsilon^{d'}(a) \mid a \in A\}$, a basis for $\tau_{B'}$. I've shown $B' \subseteq B$ and so $\tau_{B'} \subseteq \tau_B$.

For showing $\tau_B \subseteq \tau_{B'}$ I've picked some $a \in A$ and $A\cap B_\varepsilon^d(x)$, a basis element of $\tau_B$ such that $a \in A\cap B_\varepsilon^d(x)$. My goal is to find some basis element $C$ of $\tau_{B'}$ such that $a \in C \subseteq A\cap B_\varepsilon^d(x).$ If $x \in A$, then $$A\cap B_\varepsilon^d(x)=B_\varepsilon^{d'}(x)$$ and so I have my needed basis element $C$. The part where I have been having trouble is if $x \notin A$. In this case, what I have tried so far is to see if I can find some $\delta >0$ such that $$a \in B_\delta^{d'}(a) \subseteq A\cap B_\varepsilon^d(x)$$ but in searching for a delta and trying to show via set containment that the $\delta$-ball is contained in the $\varepsilon$-ball intersected with $A$ has been difficult. I've tried using triangle inequality but I am running into a problem of not knowing how to show that if $y \in B_\delta^{d'}(a)$, meaning $d'(a,y)$ that $y \in A\cap B_\varepsilon^d(x)$, mainly that $d(x,y) < \varepsilon$. It seems with the different $\delta$'s I have tried, I can use the triangle inequality to show something like $d(x,y) < \frac{3}{2}\varepsilon$ but not quite $\varepsilon$.

Suppose that $$a\in A\cap B_\epsilon^d(x)$$ for some $$x\in X$$. You know that $$\{B_r^{d'}(a):r>0\}$$ is a local base at $$a$$ in $$A$$, so there must be some $$\delta>0$$ such that $$B_\delta^{d'}(a)\subseteq B_\epsilon^d(x)$$, and hence $$B_\delta^{d'}(a)$$ is a $$d'$$-nbhd of $$a$$ contained in $$A\cap B_\epsilon^d(x)$$. Since you can do this for each $$a\in A\cap B_\epsilon^d(x)$$, $$A\cap B_\epsilon^d(x)\in\tau_{B'}$$.