Convergence of sequence extracted from a double sequence Let $X$ be a first countable topological space (e.g. any metric space).
Consider a double sequence $\{x_{n,m}\}_{m,n\in \mathbb{N}}$ on $X$ such
that $\lim_{m\rightarrow \infty }x_{n,m}$ $=$ $x_{n}$ for each $n$, and $\lim_{n\rightarrow \infty }x_{n}$ $=$ $x$. Then there exists a sequence $\{n_{i}\}_{i\in \mathbb{N}}$ tending to infinity and a sequence $m_{i}$ $=$ $m(n_{i})$ such that $\lim_{i\rightarrow \infty }x_{n_{i},m_{i}}$ $=$ $x$.
See, e.g., Lemma 1 in  Boehme and Rosenfeld (1974), who omit a proof.


*

*It seems obvious, but I am struggling to understand why the above is true.

*Clearly $\{n_{i}\}_{i\in \mathbb{N}}$ is not unique. It also seems
trivial that for any chosen valid $\{n_{i},m_{i}\}_{i\in \mathbb{N}}$, any $\{n_{i}^{\ast }\}_{i\in \mathbb{N}}$ where $n_{i}^{\ast }$ $\rightarrow $ $\infty $ as $i$ $\rightarrow $ $\infty $, and $n_{i}^{\ast }/n_{i}$ $\rightarrow $ $0$ as $i$ $\rightarrow $ $\infty $, must also satisfy $\lim_{i\rightarrow \infty }x_{n_{i}^{\ast },m(n_{i})}$ $=$ $x$. Notice the original $m(n_{i})$ is used.
This seems obvious precisely because $\lim_{m\rightarrow \infty }x_{n,m}$ $=$
$x_{n}$ and $x_{n}$ $\rightarrow $ $x$. For example, if $m(n_{i})$ $=$ $i$
and $n_{i}$ $=$ $i$ are valid, then $\lim_{i\rightarrow \infty
}x_{[i^{\delta }],i}$ $=$ $x$ must hold for any $\delta $ $\in $ $(0,1]$
where $[\cdot ]$ is the integer part.
I have scanned related topics (e.g. Extracting a convergent sequence out of a double sequence, and many others), and cannot find a suitable discussion of
this. My apologies if either (1) or (2) are painfully trivial, or (2) is
simply wrong. Help is appreciated.
Boehme, T. K.; Rosenfeld, M., An example of two compact Hausdorff Frechet spaces whose product is not Frechet, J. Lond. Math. Soc., II. Ser. 8, 339-344 (1974). ZBL0289.54026.
 A: Let $n_{i}=i$ and let $m_{i}$ be such that $|x_{n}-x_{n,m_{n}}|<\frac{1}{n}$. Note that such $m_{i}$ exists as $\lim_{m\rightarrow\infty}x_{n,m}=x_{n}$. Then
$$x_{n}-\frac{1}{n}\leq x_{n,m_{n}}\leq x_{n}+\frac{1}{n}.$$
By the algebra of limits 
$$\lim_{n\rightarrow\infty}x_{n}-\frac{1}{n}=\lim_{n\rightarrow\infty}x_{n}+\frac{1}{n}=x$$
and thus by the Sandwhich theorem (or Squeeze theorem);
$$\lim_{n\rightarrow\infty}x_{n,m_{n}}=x.$$
Your second point is not true. Consider
$$x_{n,m}=\begin{cases}1\text{ if }m\leq n^{2}\text{ if }n\text{ is not a square}\\
1\text{ if }m\leq n-1\text{ if }n\text{ is a square}\\0\text{ else}\end{cases}$$
Clearly $\lim_{m\rightarrow\infty}x_{n,m}=0=x_{n}$ and $\lim_{n\rightarrow\infty}x_{n}=0$ and if we take $n_{i}=i^{2}$ and $m_{i}=i^{2}$ then $$\lim_{i\rightarrow\infty}x_{n_{i},m_{i}}=0$$
But if we pick $n_{i}^{*}=i$ we find that $$\lim_{i\rightarrow\infty}\frac{n_{i}^{*}}{n_{i}}=\lim_{i\rightarrow\infty}\frac{1}{i}=0,$$
but 
$$x_{n_{i}^{*},m_{i}}=\begin{cases}1\text{ if }i\text{ is not suare}\\0\text{ if }i\text{ is square}\end{cases}$$
hence the sequence does not converge.
