[proofplan] The proof is a direct verification of the Lipschitz inequality for the restricted map. We take two arbitrary points of the subspace $Y$, use the defining equality between the subspace metric and the ambient metric, and then apply the Lipschitz inequality for $f$ on all of $X$. Finally, we rewrite the values of $f$ on points of $Y$ as values of the restriction $f|_Y$. [/proofplan] [step:Compare the restricted distance with the ambient distance] Let $y_1,y_2\in Y$ be arbitrary. Since $Y$ carries the subspace metric induced by $d_X$, the distance between $y_1$ and $y_2$ in $Y$ is \begin{align*} d_Y(y_1,y_2)=d_X(y_1,y_2). \end{align*} [guided] We begin with the exact pair of points for which the Lipschitz condition must be checked. Let $y_1,y_2\in Y$ be arbitrary. Because $Y$ is not being given an unrelated metric, but the subspace metric inherited from $X$, the distance function on $Y$ is precisely the restriction of $d_X$ to $Y\times Y$. Therefore \begin{align*} d_Y(y_1,y_2)=d_X(y_1,y_2). \end{align*} This identity is the only place where the subspace metric hypothesis is used. [/guided] [/step] [step:Apply the Lipschitz inequality for the ambient map] Because $f:X\to Z$ is Lipschitz with constant $L$, for all $x_1,x_2\in X$ one has \begin{align*} d_Z(f(x_1),f(x_2))\leq Ld_X(x_1,x_2). \end{align*} Since $y_1,y_2\in Y\subset X$, this inequality applies with $x_1=y_1$ and $x_2=y_2$, giving \begin{align*} d_Z(f(y_1),f(y_2))\leq Ld_X(y_1,y_2). \end{align*} Using $d_X(y_1,y_2)=d_Y(y_1,y_2)$, we obtain \begin{align*} d_Z(f(y_1),f(y_2))\leq Ld_Y(y_1,y_2). \end{align*} [/step] [step:Rewrite the inequality for the restriction] The restriction map is \begin{align*} f|_Y:Y&\to Z \end{align*} and satisfies $(f|_Y)(y)=f(y)$ for every $y\in Y$. Hence \begin{align*} d_Z((f|_Y)(y_1),(f|_Y)(y_2))\leq Ld_Y(y_1,y_2). \end{align*} Since $y_1,y_2\in Y$ were arbitrary, $f|_Y:Y\to Z$ is Lipschitz with constant $L$. [/step]