Theorem 5: If ${\displaystyle \lim _{n\to \infty }a_{m,n}=b_{m}}$ is uniformly (in $m$) and ${\displaystyle \lim _{m\to \infty }a_{m,n}=c_{n}}$ for each large $n$ then both ${\displaystyle \lim _{m\to \infty }b_m}$ and ${\displaystyle \lim _{n\to \infty }c_n}$ exist and are equal, i.e., $${\displaystyle \lim _{m\to \infty }\lim _{n\to \infty }a_{n,m}=\lim _{n\to \infty }\lim _{m\to \infty }a_{n,m}}$$
And now as corollary given that
If ${\displaystyle \sum _{n=1}^{\infty }a_{m,n}}$ converges uniformly (in m) and ${\displaystyle \sum _{m=1}^{\infty }a_{m,n}}$ converges for each large n then
$${\displaystyle \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }a_{m,n}=\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }a_{m,n}} \tag{1}$$
Proof. Direct application of Theorem 5 on $${\displaystyle S_{k,\ell }=\sum _{m=1}^{k}\sum _{n=1}^{\ell }a_{m,n}}$$
Here I want to prove using given hint. Since $${\displaystyle \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }a_{m,n}= \lim _{k\to \infty }\left \{\lim _{l\to \infty } S_{k,l}\right\}}$$
It is sufficient to prove
$${\displaystyle \lim _{k\to \infty }\lim _{l\to \infty }S_{k,l}=\lim _{l\to \infty }\lim _{k\to \infty }S_{k,l}}$$
Now if we want to use theorem 5 then we need uniform convergence of $\lim\limits_{l\to \infty} S_{k,l}$ in $k$. And $\lim\limits_{k\to \infty} S_{k,l}$ exist for each $l$. And in Hypothesis of corollary given uniform convergence of ${\displaystyle \sum _{n=1}^{\infty }a_{m,n}}$ in $m$. And ${\displaystyle \sum _{m=1}^{\infty }a_{m,n}}$ exist for each $n$. So if we prove uniform convergence of ${\displaystyle \sum _{n=1}^{\infty }a_{m,n}}$ in $m$ implies uniform convergence of $\lim\limits_{l\to \infty} S_{k,l}$ in $k$. And ${\displaystyle \sum _{m=1}^{\infty }a_{m,n}}$ exist for each $n$ implies $\lim\limits_{k\to \infty} S_{k,l}$ exist for each $l$. then we can use Theorem 5 and we are done.
Now,
define $T_{xy}={\displaystyle \sum _{n=1}^{y}a_{x,n}}$ and $f_{d} :\mathbb N\to\mathbb R$ as $m\mapsto T_{md}$, then uniform convergence of ${\displaystyle \sum _{n=1}^{\infty }a_{m,n}}$ is just unifrom convergence of $f_d ~$ on $\mathbb N$. And $g_{c} :\mathbb N\to\mathbb R$ as $k\mapsto \sum_{j=1}^{k}T_{jc}$ then uniform convergence of$~~$ $\lim\limits_{l\to \infty} S_{k,l}$ in $k$ is just unifrom convergence of $g_c ~$ on $\mathbb N$. Here $f_d$ and $g_c$ are clearly different.
So how can we show uniform convergence of $f_d$ implies uniform convergence of $g_c$?