I would love some feedback on this proof. I've spent forever working on it. At this point my mind is too jumbled to realize if I can make it more efficient. I feel confident in it, but I will note the areas I felt less sure about. Any feedback would be appreciated!
Suppose $X$ is a Banach space, and $(x_{mn})$ is a doubly indexed sequence in $X$ such that $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}||x_{mn}||<\infty.$$ Prove that $$\sum_{m=1}^{\infty}\Bigg(\sum_{n=1}^{\infty}x_{mn}\Bigg)=\sum_{n=1}^{\infty}\Bigg(\sum_{m=1}^{\infty}x_{mn}\Big).$$
We first note that $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}||x_{mn}||=\sum_{m=1}^{\infty}\Bigg(\sum_{n=1}^{\infty}||x_{mn}||\Bigg)=\sum_{n=1}^{\infty}\Bigg(\sum_{m=1}^{\infty}||x_{mn}||\Bigg),$$ which implies that \begin{align} \sum_{m=1}^M\Bigg(\sum_{n=1}^{\infty}||x_{mn}||\Bigg)&<\infty,\hspace{5 mm}\forall M\in\mathbb{N}\hspace{5 mm}(1)\\ \sum_{n=1}^N\Bigg(\sum_{m=1}^{\infty}||x_{mn}||\Bigg)&<\infty,\hspace{5 mm}\forall N\in\mathbb{N}\hspace{5 mm}(2) \end{align}
Then, for any $\epsilon>0$ we may find a $K$ so that \begin{align*} \bigg|\sum_{m=1}^{m_1}\bigg(\sum_{n=1}^{\infty}||x_{mn}||\bigg)-\sum_{m=1}^{m_2}\bigg(\sum_{n=1}^{\infty}||x_{mn}||\bigg)\bigg|&<\frac{\epsilon}{2}\\ \bigg|\sum_{n=1}^{n_1}\bigg(\sum_{m=1}^{\infty}||x_{mn}||\bigg)-\sum_{n=1}^{n_2}\bigg(\sum_{m=1}^{\infty}||x_{mn}||\bigg)\bigg|&<\frac{\epsilon}{2}. \end{align*} whenever $m_1,m_2,n_1,n_2>K$. We may use this to show that $\sum_{m=1}^M\sum_{n=1}^Nx_{mn}$ is Cauchy for any $M,N>K$. Without loss of generality, there are two cases to consider. Either $m_1>m_2$ and $n_1>n_2$ or $m_1>m_2$ and $n_2>n_1$. In the first case we see that \begin{align*} \bigg|\bigg|\sum_{m=1}^{m_1}\sum_{n=1}^{n_1}x_{mn}-\sum_{m=1}^{m_2}\sum_{n=1}^{n_2}x_{mn}\bigg|\bigg|&=\bigg|\bigg|\sum_{m=m_2+1}^{m_1}\sum_{n=1}^{n_1}x_{mn}+\sum_{m=1}^{m_1}\sum_{n=n_2+1}^{n_1}\bigg|\bigg|\\ &\leq \sum_{m=m_2+1}^{m_1}\sum_{n=1}^{n_1}||x_{mn}||+\sum_{m=1}^{m_1}\sum_{n=n_2+1}^{n_1}||x_{mn}||\\ &\leq\sum_{m=m_2+1}^{m_1}\sum_{n=1}^{\infty}||x_{mn}||+\sum_{n=n_2+1}^{n_1}\sum_{m=1}^{\infty}||x_{mn}||\\ &<\epsilon. \end{align*} While the second case gives us \begin{align*} \bigg|\bigg|\sum_{m=1}^{m_1}\sum_{n=1}^{n_1}x_{mn}-\sum_{m=1}^{m_2}\sum_{n=1}^{n_2}x_{mn}\bigg|\bigg|&=\bigg|\bigg|\sum_{m=m_2+1}^{m_1}\sum_{n=1}^{n_1}x_{mn}-\sum_{m=1}^{m_1}\sum_{n=n_1+2}^{n_2}x_{mn}\bigg|\bigg|\\ &\leq \sum_{m=m_2+1}^{m_1}\sum_{n=1}^{n_1}||x_{mn}||+\sum_{n=n_1+2}^{n_2}\sum_{m=1}^{m_1}||x_{mn}||\\ &\leq \sum_{m=m_2+1}^{m_1}\sum_{n=1}^{\infty}||x_{mn}||+||\sum_{n=n_1+2}^{n_2}\sum_{m=1}^{\infty}||x_{mn}||\\ &<\epsilon. \end{align*} Since this inequality arises with $m_1,m_2,n_1,n_2$ chosen independently, we are able to take their limits independently in order to arrive at the desired equality. This is made rigorous by the following argument (the following is an area of relative uncertainty for me. I believe it is valid but was somewhat not confident). Given $\epsilon>0$ we find $K_1$ so that $$\bigg|\bigg|\sum_{m=1}^{m_1}\sum_{n=1}^{n_1}x_{mn}-\sum_{m=1}^{m_2}\sum_{n=1}^{n_2}x_{mn}\bigg|\bigg|<\frac{\epsilon}{2}$$ whenever $m_1,m_2,n_1,n_2>K_1$. We then find $K_2$ such that $$\bigg|\bigg|\sum_{m=1}^{m_2}\sum_{n=1}^{\infty}x_{mn}-\sum_{m=1}^{m_2}\sum_{n=1}^{n_2}x_{mn}\bigg|\bigg|<\frac{\epsilon}{2}$$ whenever $n_2>K_2$, which is guaranteed by $(1)$. Then, letting $K>\max(K_1,K_2)$, it follows that \begin{align*} \bigg|\bigg|\sum_{m=1}^{m_1}\sum_{n=1}^{n_1}x_{mn}-\sum_{m=1}^{m_2}\sum_{n=1}^{\infty}x_{mn}\bigg|\bigg|&\leq \bigg|\bigg|\sum_{m=1}^{m_1}\sum_{n=1}^{n_1}x_{mn}-\sum_{m=1}^{m_2}\sum_{n=1}^{n_2}x_{mn}\bigg|\bigg|\\ &+\bigg|\bigg|\sum_{m=1}^{m_2}\sum_{n=1}^{\infty}x_{mn}-\sum_{m=1}^{m_2}\sum_{n=1}^{n_2}x_{mn}\bigg|\bigg|\\ <\epsilon \end{align*} whenever $m_1,m_2,n_1>K$. We may repeat this process to arrive at $$\bigg|\bigg|\sum_{m=1}^{\infty}\bigg(\sum_{n=1}^{\infty}x_{mn}\bigg)-\sum_{n=1}^{\infty}\bigg(\sum_{m=1}^{\infty}x_{mn}\bigg)\bigg|\bigg|<\epsilon$$ for all $\epsilon$.
To be clear at the onset, the convergence of a double series $\sum_{m,n} x_{mn} = \sum_m \sum_n x_{mn}$ means there exists $S$ such that for any $\epsilon > 0$, there exists $N(\epsilon) \in\mathbb{N}$ such that $\|\sum_{m=1}^M\sum_{n=1}^N x_{mn} - S\| < \epsilon$ for all $m, n > N(\epsilon)$. The convergence of a double series does not necessarily follow from the convergence of an iterated series $\sum_{m=1}^\infty\left(\sum_{n=1}^\infty x_{mn}\right)$ and vice versa.
With this definition, it is true that convergence of the double series $\sum_{m=1}^\infty\sum_{n=1}^\infty \|x_{mn}\|$ implies convergence of the double series $\sum_{m=1}^\infty\sum_{n=1}^\infty x_{mn}$ and equality of the iterated series
$$\sum_{m=1}^\infty\sum_{n=1}^\infty x_{mn} = \sum_{m=1}^\infty\left(\sum_{n=1}^\infty x_{mn}\right) = \sum_{n=1}^\infty\left(\sum_{m=1}^\infty x_{mn}\right)$$
This can be proved in a series of steps, some of which (but not all) you did correctly and/or clearly.
(1) As you stated but did not prove,
$$\sum_{m=1}^\infty\sum_{n=1}^\infty \|x_{mn}\| = \sum_{m=1}^\infty\left(\sum_{n=1}^\infty \|x_{mn}\|\right) = \sum_{n=1}^\infty\left(\sum_{m=1}^\infty \|x_{mn}\|\right),$$
holds when the double series on the LHS converges. This follows from the nonnegativity of $\|x_{mn}\|$ and a straightforward monotone convergence argument.
(2) It then follows as you correctly showed that the double series $\sum_{m,n} x_{mn}$ is convergent by the Cauchy criterion, with
$$\sum_{m=1}^\infty\sum_{n=1}^\infty x_{mn} = S$$
(3) The series $\sum_{n=1}^\infty x_{mn}$ and $\sum_{m=1}^\infty x_{mn}$ converge for each $m$ and for each $n$, respectively. This follows again by a Cauchy argument using, for example,
$$\left\|\sum_{n= n_1+1}^{n_2} x_{mn} \right\| \leqslant \sum_{n= n_1+1}^{n_2} \|x_{mn}\|$$
We can finish (easily) as follows. From convergence established in (2), for any $\epsilon > 0$, there exists $K(\epsilon) \in \mathbb{N}$ such that for all $M,N > K(\epsilon)$,
$$\|S_{MN} - S\| := \left\|\sum_{m=1}^M\sum_{n=1}^N x_{mn} -S\right\| < \epsilon $$
From (3) there exists $T_M$ for all $M > K(\epsilon)$ such that
$$T_M = \lim_{N \to \infty}S_{MN} = \sum_{m=1}^M\left(\sum_{n=1}^\infty x_{mn}\right)$$
Finally, we have for all $M > K(\epsilon)$,
$$\left\|\sum_{m=1}^M\left(\sum_{n=1}^\infty x_{mn}\right) -S\right\|= \|\lim_{N \to \infty} S_{MN} - S \| = \lim_{N \to \infty} \|S_{MN} -S\| \leqslant \epsilon,$$
which implies that
$$\sum_{m=1}^\infty\left(\sum_{n=1}^\infty x_{mn}\right) =S = \sum_{m=1}^\infty\sum_{n=1}^\infty x_{mn} $$
The same result can be proved for the other iterated series in a similar way.
The key step in the last part of the proof is the following general result. For any convergent sequence $a_n$ in a Banach space such that $\|a_n - C\| < \epsilon$ for all sufficiently large $n$, it follows that $\|\lim_{n \to \infty}a_n - C\|= \lim_{n \to \infty}\|a_n - C\|\leqslant \epsilon$.
For proof, let $\lim_{n \to \infty} a_n = L$. By the reverse triangle inequality,
$$|\, \|a_n - C\| - \|L - C\|\, |\leqslant \|(a_n - C)- (L-C) \| = \|a_n - L\| \underset{n \to \infty}\longrightarrow 0$$
Hence, $\lim_{n \to \infty} \|a_n -C\| = \|L-c\| = \|\lim_{n\to \infty}a_n -C\|$.
Given that $\|a_n - C\| < \epsilon$ for all $n > N_1$, assume that $\|L-C\| > \epsilon$. Take $\eta = (\|L-C\|+\epsilon)/2$. Since $\epsilon < \eta < \|L-C\|$, there exists $N_2$ such that for all $n > N_2$ we have
$$\|a_n - C\| > \eta > \epsilon,$$
which is a contradiction for $n > \max (N_1,N_2)$.
Therefore, $\|\lim_{n \to \infty} a_n - c\|= \lim_{n \to \infty}\|a_n - C\| \leqslant \epsilon$.