Iterated Limits Proof If $\lim\limits_{(x,y) \to (a,b)} f(x,y) = L$, and if the one-dimensional limits
$~\lim\limits_{x \to a} f(x,y), ~\lim\limits_{y \to b} f(x,y)~$ both exist, prove that
$$\lim\limits_{x \to a} \Bigg(\lim\limits_{y \to b} f(x,y)\Bigg) = \lim\limits_{y \to b} \Bigg(\lim\limits_{x \to a} f(x,y)\Bigg) = L$$
Proof. Since  both one-dimensional limits exist, let $~\lim\limits_{x \to a} f(x,y) = L_a, ~ \lim\limits_{y \to b} f(x,y) = L_b$.
For every $\epsilon > 0$ there exist $\delta > 0$ such that if the points $(x,y)$ are in the neighborhood $V_\delta (a,b)$, the points $f(x,y)$ are in the neighborhood $V_\epsilon (f(a,b))$, in other words
$$0 < |(x,y) - (a,b)| < \delta$$ implies $$|f(x,y) - L| < \epsilon/2$$
Also, for every $\epsilon > 0$ there exist $\delta > 0$ such that if $0 < |x-a| < \delta$, then $|f(x,y) - L_a| < \epsilon/2$
From the properties of inequality, $|x| - |y| \leq |x-y|$,
$$|L_a - L| - |L_a - f(x,y)| \leq |f(x,y) - L|$$
$$|L_a - L| \leq |f(x,y) - L| + |L_a - f(x,y)| < \epsilon/2 + \epsilon/2 = \epsilon$$
Thus, for every $\epsilon > 0$ there exist $\delta > 0$ such that if $0 < |y-b| < \delta$, then $|L_a - L| < \epsilon$. This implies that,
$$\lim\limits_{x \to a} \Bigg(\lim\limits_{y \to b} f(x,y)\Bigg) = L$$
By appyling the same argument with the other limit, we conclude that 
$$\lim\limits_{y \to b} \Bigg(\lim\limits_{x \to a} f(x,y)\Bigg) = L$$
Can anyone comment on my work? Have I missed something? 
 A: Insofar as $L_a$ and $L_b$ are constants, the proof is acceptable. However, the hypothesis is that $\lim_{x \to a} f(x,y)$ and $\lim_{y \to b} f(x,y)$ exist. This  means the limits may be non-constant functions $x \mapsto L_b(x)$ and  $y \mapsto L_a(y)$.  
Your approach is still generally correct, however to be rigorous you must specify the "deltas" carefully.
Since the double limit exists we can specify your neighborhood $V_{\delta}(a,b)$ or more directly claim there exists $\delta_1(\epsilon,a,b)$ such that if $|x-a|,|y-b| < \delta_1(\epsilon,a,b)$ then $|f(x,y) - L| < \epsilon/2$.
Since $\lim_{x \to a} f(x,y) = L_a(y)$ there exists $\delta_2(\epsilon,y)$, which may depend on $y$, such that if $|x-a| < \delta_2(\epsilon,y)$ then $|f(x,y) - L_a(y)| < \epsilon/2$ for each $y$ in the domain (left unspecified by you).
Suppose $|y - b| < \delta_1(\epsilon, a,b)$. We can chose $\hat{x}(y)$ (which may depend on $y$) such that
$$|\hat{x}(y) - a| < \min(\delta_1(\epsilon,a,b), \delta_2(\epsilon,y)), $$
and $|f(\hat{x}(y),y) - L_a(y)|, \,  |f(\hat{x}(y),y) - L| < \epsilon/2.$ 
Proceeding, as you did, it follows that if $|y-b| < \delta_1(\epsilon,a,b)$, then
$$|L_a(y) - L| \leqslant |f(\hat{x}(y),y) - L| + |f(\hat{x}(y),y) - L_a(y)| < \epsilon/2 + \epsilon/2 =  \epsilon,$$ 
proving
$$\lim_{y \to b} \left(\lim_{x \to a} f(x,y) \right) = \lim_{y \to b} L_a(y) =L$$
Another Proof
A less cumbersome argument is that there exists $\delta(\epsilon)$ such that when $|x-a|, \, |y-b| < \delta(\epsilon)$, we have
$$|f(x,y) - L| < \epsilon.$$ 
Hence,
$$| \lim_{x \to a}f(x,y) - L| = \lim_{x \to a}|f(x,y) - L| \leqslant \epsilon.$$
This uses
$$g(x) < \epsilon \implies \lim_{x \to a} g(x) \leqslant \epsilon,$$
and
$$| \, |f(x,y) - L| - |\lim_{x \to a}f(x,y)-L| \, | \leqslant |f(x,y) - \lim_{x \to a} f(x,y)|$$
A: As RRL has noted you have tacitly assumed that $L(x):=\lim_{y\to b} f(x,y)$ is constant. This assumption is not in the books, and is unnecessary.
In order to streamline the argument I'm assuming that $a=b=0$ and that $f(0,0):=L=0$.
Given an $\epsilon>0$ there is a $\delta>0$ such that
$$|x|\leq \delta \ \  \wedge \ \  |y|\leq \delta\quad\Longrightarrow \quad\bigl|f(x,y)\bigr|\leq\epsilon\ .$$
Fix an $x$ with $|x|\leq\delta$. Then $\bigl|f(x,y)\bigr|\leq\epsilon$ for all $y\in[{-\delta},\delta]$ and therefore $$L(x):=\lim_{y\to0}f(x,y)\in [{-\epsilon},\epsilon]$$
as well. Since this is true for all $x\in[{-\delta},\delta]$, and $\epsilon>0$ was arbitrary,  we can conclude that $\lim_{x\to0} L(x)=0=L$.
