Suppose $X$ is a metric space, and for all positive integers $m$ and $n$, we have an element $s_{m,n}\in X$ such that $\lim_{n \to \infty}s_{m,n}$ exists, and such that $\lim_{m \to \infty}\lim_{n\to \infty}s_{m,n}=p.$ Show that there exists a sequence of positive integers $N_1,N_2,...$ such that for every sequence of positive integers $n_1,n_2,...$ satisfying $n_m\geq N_m$ one has $\lim_{m \to \infty} s_{m,n_m}=p$.
Is the general idea of my proof correct? I feel like I must be missing something. Here is my attempt:
Let $\lim_{n\to \infty}s_{m,n}=p_m.$ Since $p_m\to p,$ then for all $\varepsilon >0$, there exists $M$ so that for $m\geq M, d(p_m,p)<\varepsilon$.
For each $m,$ we have $s_{m,n}\to p_m$. Then we can choose $N_m$ so that for all $r\geq N_m$, $d(s_{m,r},p_m)<\frac{1}{m}.$
Fix $m$. Then for every sequence $n_1,n_2,...$ satisfying $n_m\geq N_m,$ we have $d(s_{m,n_m},p)\leq d(s_{m,n_m},p_m)+d(p_m,p)<\frac{1}{m}+d(p_m,p).$
Taking $m$ to infinity, we have $d(s_{m,n_m},p)<0+\varepsilon=\varepsilon$
You're definitely on the right track. Choose $\varepsilon \gt 0$. Using your definition of $p_m$, we know that $\exists M$ such that $m \gt M \Rightarrow \vert p - p_m \vert \lt \frac{\varepsilon}{2}$. We also know for each $m \exists N_m$ such that $n \gt N_m \Rightarrow \vert p_m - s_{m, n} \vert \lt \frac{\varepsilon}{2}$.
Then by the triangle inequality, $m \gt M \land n \gt N_m \Rightarrow \vert p - s_{m, n} \vert \lt \varepsilon$. Since $\varepsilon \gt 0$ was arbitrary and $n_m \gt N_m$ was also arbitrary, this means $\lim_{m \to \infty} s_{m, n_m} = p$.
This just makes more formal what you're doing when you "[t]ak[e] $m$ to infinity."