Let $f \colon \mathbb{R}^2 \to \mathbb{R}$ be a function, let $(a, b) \in \mathbb{R}^2$, and let $L \in \mathbb{R}$. Suppose that $\lim_{ (x,y) \to (a, b) } f(x, y)$ exists and equals $L$, and suppose also that the one-dimensional limits $\lim_{x \to a} f(x, y)$ and $\lim_{y \to b} f(x, y)$ both exist. Then how to prove that the two iterated limits $\lim_{x \to a} \left[ \lim_{y \to b} f(x, y) \right]$ and $\lim_{ y \to b } \left[ \lim_{ x \to a } f(x, y) \right]$ both exist and that we have the following result? $$ \lim_{x \to a} \left[ \lim_{y \to b} f(x, y) \right] = \lim_{ y \to b} \left[ \lim_{ x \to a } f(x, y) \right] = L.$$
My Attempt:
Let $\varepsilon > 0$ be given. Then we can find a real number $\delta > 0$ such that $$ \left\lvert f(x, y) - L \right\rvert < \varepsilon $$ for all $(x, y) \in \mathbb{R}^2$ for which $$ 0 < \sqrt{ (x-a)^2 + (y-b)^2 } < \delta. $$
Let $\phi \colon \mathbb{R} \to \mathbb{R}$ and $\psi \colon \mathbb{R} \to \mathbb{R}$ be the functions defined by $$ \phi(x) = \lim_{y \to b} f(x, y) \ \mbox{ for all } \ x \in \mathbb{R}, $$ and $$\psi(y) = \lim_{x \to a} f(x, y) \ \mbox{ for all } \ y \in \mathbb{R}. $$
Now we need to show that $\lim_{x \to a } \phi(x)$ and $\lim_{y \to b} \psi(y)$ both exist and that $$ \lim_{x \to a } \phi(x) = L = \lim_{y \to b} \psi(y). $$
How to proceed from here?
The beginning is OK. Now, in order to prove by definition that
$$\lim_{x \to a} \phi(x) = L,$$
it suffices to show that the neighborhood $|x-a| < \delta$ is correct for the chosen $\varepsilon > 0$.
Fix $x_0$ such that $|x_0 - a| < \delta$. Then there is a $\delta' > 0$ such that
$$\{ x_0 \} \times (b - \delta', b + \delta') \subseteq B( (a, b), \delta),$$
so in the neighborhood $(b-\delta', b+\delta')$ of $b$ we have $|f(x_0, y) - L| < \varepsilon$, hence $|\phi(x_0) - L| \leqslant \varepsilon$.
The second part is done in a symmetrical fashion.