Is the Following argument correct?
Given a doubly indexed array $a_{mn}$ where $m,n\in\mathbf{N}$. Assume that $\lim_{m,n\to\infty}a_{mn} = a$, and given that for each fixed $m\in\mathbf{N}$, $\lim_{n\to\infty}(a_{mn}) = b_{m}$. Show that $\lim_{m\to\infty}b_m = a$.
Proof. Let $\epsilon>0$, since $\lim_{m,n\to\infty}a_{mn} = a$ there exists an $M\in\mathbf{N}$ such that $\forall m,n\ge M\left(|a_{mn}-a|<\frac{\epsilon}{2}\right)$. Now let $m\ge M$, from hypothesis $\lim_{n\to\infty}a_{mn} = b_m$ and thus for some $N\in\mathbf{N}$ we have $|a_{mn}-b_m|<\frac{\epsilon}{2}$ whenever $n\ge N$.
Now let $L = \max(M,N)$ and $n\ge L$ then we have $|b_m-a_{mn}|<\frac{\epsilon}{2}$ and $|a_{mn}-a|<\frac{\epsilon}{2}$, appealing to the triangle inequality we have $$|b_m-a| = |(b_m-a_{mn})+(a_{mn}-a)|\leq|b_{m}-a_{mn}|+|a_{mn}-a|<\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$$
$\blacksquare$
PS: We define $\lim_{m,n\to \infty}(a_{mn})=a$ if and only if $\forall\epsilon>0\exists M\in\mathbf{N}\forall m,n\ge M(|a_{mn}-a|<\epsilon)$.
Your proof is only valid under the assumption that $a_{mn} \to b_m$ uniformly for $m \in \mathbb{N}$.
To fix this, note that by the reverse triangle inequality, we have
$$0 \leqslant | \,|a_{mn} - a | - | b_m - a| \, | \leqslant |a_{mn} - b_m| $$ $$\implies 0 \leqslant \lim_{n \to \infty}| \,|a_{mn} - a | - | b_m - a| \, | \leqslant \lim_{n \to \infty}|a_{mn} - b_m| = 0 \\ \implies \lim_{n \to \infty}|a_{mn} - a| = |b_m - a|$$
SInce $|a_{mn} - a| < \epsilon $ for all $m,n > M$, we have for all $m > M$
$$|b_m - a| = \lim_{n \to \infty}|a_{mn} -a | \leqslant \epsilon,$$
proving $\lim_{m \to \infty} b_m = a$.