Exercise 1.1.16 (Tao, 2011) Let $(X,d)$ be a metric space. Let $(x_n)_{n=n_0}^{\infty}$ and $(y_n)_{n=n_0}^{\infty}$ be two sequences in $X$ such that $(x_n)_{n=n_0}^{\infty}$ converges to $x\in X$ and $(x_n)_{n=n_0}^{\infty}$ converges to $y \in X$ w.r.t. metric $d$. Show that $$lim_{n\to \infty} d(x^n,y^n) = d(x,y)$$
Proof: using the defintion of convergence in metric spaces we have to show that $$\forall \epsilon > 0 \quad \exists n_{\epsilon} \quad \forall n \geq n_{\epsilon}: \quad d(d(x^n,y^n),d(x,y)) \leq \epsilon$$ I have shown using triangle inequality that $$d(x^n,y^n) \leq d(x^n,y) + d(y,y^n) \leq d(x^n,x) + d(x,y) + d(y,y^n) \leq 2\epsilon + d(x,y)$$ Thus $$d(x^n,y^n) - d(x,y)\leq 2\epsilon$$ But I cannot use this fact to conclude that $$d(d(x^n,y^n),d(x,y)) \leq 2\epsilon$$ How do I proceed then?