Just need to check the details in my proof are correct
Question
$\displaystyle (X,d)$ is a metric space, fix $\displaystyle x_0\in{X}$, let $\displaystyle F_{x_0}=d(x_0,x)$. Show this function is continuous

Proof
Consider the open set $\displaystyle W\in{\mathbb{R}}$ and let
$\displaystyle F_{x_0}^{-1}(W)=[x\in{X}|F_{x_0}(x)=d(x_0,x)\in{W}}$]. For all $\displaystyle y\in{W}$, $\displaystyle \exists{\epsilon_y}>0$ such that $\displaystyle B(y,\epsilon_y)\subset(W)$
Let $\displaystyle z=F_{x_0}^{-1}$ so $\displaystyle z\in{F_{x_0}^{-1}}$, consider $\displaystyle B(z,\epsilon_y)$

Let $\displaystyle \alpha{\in{B(z,\epsilon_y)}$,$\displaystyle F_{x_0}^{-1}(\alpha)$$\displaystyle =d(x_0,\alpha)<d(x_0,z)+d(z,\alpha)=y+\epsilon_y$
So $\displaystyle F_{x_0}^{-1}(\alpha)\in{B(y,\epsilon_y)}\subset{W}$, so $\displaystyle B(z,\epsilon_y)\subset{F_{x_0}^{-1}(W)}$
So $\displaystyle {F_{x_0}^{-1}(W)$ is open and the result follows