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

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

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