Given metric spaces $(X,\rho_X)$ and $(Y,\rho_Y)$, and a function $f : X \to Y$, we say that $f$ is continuous with respect to $\rho_X$ and $\rho_Y$ (in shorthand, $f : (X,\rho_X) \to (Y,\rho_Y)$ is continuous) if for any $x_0 \in X$ and any $\varepsilon>0$ there exists $\delta>0$ (that depends on $x_0$ and $\varepsilon$) such that $$(\forall x \in X) \quad \rho_X(x,x_0)<\delta \!\implies\! \rho_Y(f(x),f(x_0)) < \varepsilon$$ So, you need to prove that $d : (A \times A,d_1) \to (\mathbb R,|\cdot|)$ is continuous, that is, given $(x_0,y_0) \in A \times A$ and $\varepsilon>0$, you need to find $\delta>0$ such that $$(\forall x,y \in A) \quad d_1((x,y),(x_0,y_0)) < \delta \!\implies\! |d(x,y)-d(x_0,y_0)| < \varepsilon.$$

Given metric spaces $(X,\rho_X)$ and $(Y,\rho_Y)$, and a function $f : X \to Y$, we say that $f$ is continuous with respect to $\rho_X$ and $\rho_Y$ (in shorthand, $f : (X,\rho_X) \to (Y,\rho_Y)$ is continuous) if for any $x_0 \in X$ and any $\varepsilon>0$ there exists $\delta>0$ (that depends on $x_0$ and $\varepsilon$) such that $$(\forall x \in X) \quad \rho_X(x,x_0)<\delta \!\implies\! \rho_Y(f(x),f(x_0)) < \varepsilon.$$ So, you need to prove that if $(A,d)$ is a metric space, then $d : (A \times A,d_1) \to (\mathbb R,|\cdot|)$ is continuous, that is, given $(x_0,y_0) \in A \times A$ and $\varepsilon>0$, you need to find $\delta>0$ such that $$(\forall x,y \in A) \quad d_1((x,y),(x_0,y_0)) < \delta \!\implies\! |d(x,y)-d(x_0,y_0)| < \varepsilon.$$