Let $ (X,d_x) $ and $(Y,d_y)$ be metric spaces, an isometry $f:(x,d_x)\rightarrow (y,d_y)$ Is Always continous
Proof:
A function Is an isometry if keeps distances unaltered therfore $ \forall x_1,x_2\in X$ $ d_x(x_1,x_2)=d_y(f(x_1),f(X_2))$
And a function is continuous when the retro image of an open set is always open, in the case of metric spaces, balls are open sets ( we can create a base of balls) therefore is enough to prove that for each ball of Y , it is contained in a ball of x
Let's consider $y_0\in Y$ and $B_y (y_0, \epsilon):=${$y| d_y(y,y_0) $}
$\forall y\in B_y (y_0, \epsilon)$
$ d_y( y,y_0)<\epsilon \implies$ $ d_x(f^{-1}(y),f{-1}(y_0)< \epsilon$
Therefore every ball of Y is also contained in a ball of X, the only problem I have is that I think that it is not correct to take the inverse function of $y_0,y$ therefore I think the proof is wrong.