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

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?