If I wanted to show that an isometry is always continuous, is this right?

Question

So I need to show that an isometry is always continuous, (from $M$ to $N$ in this case) and my first thought was to show that for some $p,q\in M$, $\exists$ $\varepsilon\gt0$ such that $d_M(p,q)\lt \varepsilon$, and for some $f(p),f(q)\in N$, $d_N(f(p),f(q))$ must also be less than epsilon. But then I realize that's not enough, because if an isometry preserves distance, then if $d_M(p,q)=\alpha$ for some $\alpha\in M \implies \alpha\in N$ and therefore $d_N(f(p),f(q))$ must also be equal to $\alpha$. Is this reasoning right? Is this even in the right direction of a proof?

Answer 1

Note: I think you meant $\alpha\in\mathbb{R}$ and not $\alpha\in M$, and anyway that statement is unclear.

Your general reasoning is correct, but the proof is vague. To show continuity do this: Let $y\in Im(f)$, and let $x \in M$ with $f(x)=y$. Now let $\epsilon>0$. We need to show that there's a $\delta > 0$ such that for every $z\in B(x,\delta)$ (the open ball of size $\delta$ around $x$), it holds that $f(z)\in B(y,\epsilon)$.

Can you see what $\delta$ should be?