Let $d'\colon Y\colon Y\to \sR$ be the restriction of $d$ to $Y$ , and let \begin{eqnarray*} B_r(x) &=& \{ z\in X: d'(z,x)<r \}, \\ B'_r(x) &=& \{ z\in Y: d'(z,x)<r\}. \end{eqnarray*}The proof rests on the identity $$ B'_r(x)=Y\cap B_r(x), \quad x\in Y, r>0. $$ Suppose $A\subseteq Y$ is open in the subspace topology of $Y$ , then $A=Y\cap V$ for some open $V\subseteq X$ . Since $V$ is open in $X$ , $$ V=\cup \{ B_{r_i}(x_i) : i=1,2, \ldots \} $$ for some $r_i>0$ , $x_i\in X$ , and \begin{eqnarray*} A&=&\cup \{ Y\cap B_{r_i}(x_i) : i=1,2, \ldots \} \\ &=&\cup \{ B'_{r_i}(x_i) : i=1,2, \ldots \}. \end{eqnarray*}Thus $A$ is open also in the metric topology of $d'$ . The converse direction is proven similarly.