Let $\{F(k,n)\,:\:, n\geq 1, m\geq1\}$ be a double sequence family of extended real numbers $\Bbb R\cup\{-\infty,\infty\}$. I would like to prove the following statements
1- There is $(n_k)_k$, $n_k\to\infty$ as $k\to\infty$ such that
$$\limsup_{k\to \infty} F(k, n_k) \leq \limsup_{n\to \infty}\limsup_{k\to \infty} F(k,n). $$
2-There is $(n_k)_k$, $n_k\to\infty$ as $k\to\infty$ such that
$$ \liminf_{k\to \infty} F(k, n_k) \leq \limsup_{n\to \infty}\liminf_{k\to \infty} F(k,n). $$
I am also looking for a good reference for these statements.
I'm just going to prove the first statement, the same arguments can be used for the second one (see edit for a proof of the second statement). Let's first assume that the right-hand side of the inequality is $<+\infty$, otherwise we directly have the inequality. For now, let's also assume it's $>-\infty$.
If $\,\boldsymbol{-\infty<\limsup\limits_{n\to\infty}\limsup\limits_{k\to\infty}F(k,n)<+\infty}$:
Since it is $<+\infty$, we have $\limsup_{k\to\infty}F(k,n)<+\infty$ for large enough $n$. Furthermore, since we assumed it's also $>-\infty$, that means there's an infinite and increasing sequence $(n(i))_{i\in\mathbb{N}}$ such that: $$ -\infty<\limsup_{k\to\infty}F(k,n(i))<+\infty \ \ \ \ \ \text{ for all } i\geq1. $$ In particular, $n(i)\underset{i\to\infty}{\longrightarrow} \infty$.
Now that we've ensured this $\limsup$ is finite, let's define a sequence $(k_i)_i$ recursively. Let's just choose $k_0=1$, and for every $i\geq1$, we define $k_i\in \mathbb{N}$ such that $k_i > k_{i-1}$ and: $$ F(k,n(i)) < \limsup_{k'\to\infty} F(k',n(i)) + \frac{1}{n(i)} \ \ \ \ \ \text{ for all } k\geq k_i. $$ We can now define our $(n_k)_k$ as follows: $$ n_k := \begin{cases} 1 &\text{if } 1\leq k < k_1,\\ n(1) &\text{if } k_1\leq k < k_2,\\ n(2) &\text{if } k_2\leq k < k_3,\\ n(3) &\text{if } k_3\leq k < k_4,\\ \vdots &\vdots \end{cases} $$ We can see that $\lim\limits_{k\to\infty}n_k = \infty$ and, for all $k\geq k_1$: $$ F(k,n_k) - \limsup_{k'\to\infty} F(k',n_k) < \frac{1}{n_k}, $$ and so we can write : \begin{align*} \limsup_{k\to\infty} F(k,n_k) &= \limsup_{k\to\infty} \left(F(k,n_k) - \limsup_{k'\to\infty} F(k',n_k) + \limsup_{k'\to\infty} F(k',n_k)\right) \\ &\leq \limsup_{k\to\infty} \underbrace{\left(F(k,n_k) - \limsup_{k'\to\infty} F(k',n_k) \right)}_{\leq 1/n_k} + \limsup_{k\to\infty} \limsup_{k'\to\infty} F(k',n_k) \\ &\leq \limsup_{n\to\infty} \limsup_{k\to\infty} F(k,n). \end{align*} Statement 1 is thus proven when the right-hand side of the inequality is $>-\infty$.
If $\,\boldsymbol{\limsup\limits_{n\to\infty}\limsup\limits_{k\to\infty}F(k,n)=-\infty}$:
When it is $-\infty$, we can instead define $(n(i))_i$ in a similar way to $(k_i)_i$, such that $n(i)>n(i-1)$ and: $$ \limsup_{k\to\infty} F(k,n(i)) < -i, $$ and similarly, define $(k_i)_i$ such that $k_i>k_{i-1}$ and: $$ F(k,n(i)) < -i \ \ \ \ \text{ for all } k\geq k_i, $$ and use the same definition for $(n_k)_k$. We still have $\lim\limits_{k\to\infty} n_k = \infty$, and: $$ \limsup_{k\to\infty} F(k,n_k) = -\infty, $$ and so the inequality is still true.
$ $
EDIT: For the second statement, we can break it down into the same two cases, since we directly have the inequality when the right-hand side is $+\infty$.
If $\,\boldsymbol{-\infty<\limsup\limits_{n\to\infty}\liminf\limits_{k\to\infty}F(k,n)<+\infty}$:
We can define $(n(i))_{i\in\mathbb{N}}$ almost like before, as a subsequence of the integers such that : $$ -\infty<\liminf_{k\to\infty}F(k,n(i))<+\infty \ \ \ \ \ \text{ for all } i\geq1. $$ Now let's instead define $(k_i)_i$ as $k_0=1$, and for every $i\geq1$, $k_i$ is such that $k_i>k_{i-1}$ and : $$ |\,F(k_i,n(i)) - \liminf_{k'\to\infty} F(k',n(i))\,| < \frac{1}{n(i)}. $$ Reusing the same definition of $(n_k)_k$, we get: $$ |\,F(k_i,n_{k_i}) - \liminf_{k'\to\infty} F(k',n_{k_i})\,| < \frac{1}{n_{k_i}} \ \ \ \ \ \text{ for all } i\geq1, $$ and we can write: \begin{align*} \liminf_{k\to\infty} F(k,n_k) &\leq \liminf_{i\to\infty} F(k_i,n_{k_i}) \\ &\leq \liminf_{i\to\infty} \left(\vphantom{\limsup_k}\right. \underbrace{F(k_i,n_{k_i}) - \liminf_{k'\to\infty} F(k',n_{k_i})}_{\underset{i\to\infty}{\longrightarrow}\, 0} + \liminf_{k'\to\infty} F(k',n_{k_i}) \left.\vphantom{\limsup_k}\right) \\ &\leq \liminf_{i\to\infty} \liminf_{k'\to\infty} F(k',n_{k_i}) \leq \limsup_{i\to\infty} \liminf_{k'\to\infty} F(k',n_{k_i}) \\ &\leq \limsup_{n\to\infty} \liminf_{k\to\infty} F(k,n). \end{align*}
If $\,\boldsymbol{\limsup\limits_{n\to\infty}\liminf\limits_{k\to\infty}F(k,n)=-\infty}$:
Almost like before, define $(n(i))_i$ such that $n(i)>n(i-1)$ and: $$ \liminf_{k\to\infty} F(k,n(i)) < -i, $$ and define $(k_i)_i$ such that $k_i>k_{i-1}$ and: $$ F(k_i,n(i)) < -i. $$ Reusing the same $(n_k)_k$ again, we have $n_{k_i} = n(i)$ and: $$ \liminf_{k\to\infty} F(k,n_k) \leq \liminf_{i\to\infty} F(k_i,n_{k_i}) = -\infty. $$