Prove that any subset of $\mathbb{R}^2$ which is $D$-open is also $D_1$-open.

Prove that any subset of $\mathbb{R}^2$ which is $D$-open is also $D_1$-open.

Prove that any subset of $\mathbb{R}^2$ which is $D$-open is $D_1$-open
and, conversely, that any subset of the plane which is $D_1$-open is
$D$-open.
Let $D$ be $D(x_1,y_1)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$. Let $D_1$ be
$D((x_1,y_1)(x_2,y_2))=\lvert x_1 - x_2 \rvert + \lvert y_1-y_2 \rvert$.
Attempt:
Consider the metric space $(X,D)$. Let $U\subset X$. Then, for any point
$(x,y)\in U$ there exists $p>0$ where $N_p(x,y)\subset U$.
Consider the metric space $(X,D_1)$. Let $V\subset X$. Then, for any point
$(x_1,y_1)\in V$ there exists $q>0$ where $N_q(x_1,y_1)\subset V$.
Given $p$, let $q=p/\sqrt{2}$. Then for any $(x',y')\in N_q(x)\subset V$,
$(x',y')\in N_p(x)\subset U$. Hence, if $X$ is $D$-open, then it must also
be $D_1$-open.
Any help fixing this proof is appreciated.