Problem Statement: Let $x,y\in\mathbb{R}^n.$ Show that for any sequences $\{x_k\}, \{y_k\}$ such that $x_k\to x$ and $y_k\to y,$ we have $$\lim_{k\to\infty}\rho(x_k,y_k)=\rho(x,y)$$ where $\rho:\mathbb{R}^{2n}\to\mathbb{R}$ is the Euclidean metric.
My Work So Far: The first solution I did was applying the Triangular Inequality theory for:
$d(x,y) \le d(x,x_k) + d(x_k,y_k) + d( y_k,y)$ and then takes a limit on both sides
Another solution was to use the limit or convergence definition: $| \|x_k-y_k\|-\|x-y\|| < \varepsilon,$ then take $\varepsilon_1=\varepsilon/2.$
Not sure/confident about both methods.
You've got a good start. I would also apply the triangle inequality the other way: \begin{align*} \rho(x_k,y_k) &\le \rho(x_k,x)+\rho(x,y_k)\\ &\le \rho(x_k,x)+\rho(x,y)+\rho(y,y_k). \end{align*} If you take the limits on both sides, you'll see that the first and third terms on the RHS go to zero. Combine this with the way you applied the triangle inequality, and you can get \begin{align*} \lim_{k\to\infty}\rho(x_k,y_k)&\le \rho(x,y)\quad\text{and}\\ \rho(x,y)&\le \lim_{k\to\infty}\rho(x_k,y_k). \end{align*}