real analysis - Is my proof of the following theorem correct?

Theorem

Let $f : \mathbb{R}^2 \to \mathbb{R}$ and $(a,b)\in \mathbb{R}^2$. Let $f_b : \mathbb{R}\to\mathbb{R}$ be defined by $f_b(x)=(x,b)$ for all $x\in \mathbb{R}$. Then, $$\displaystyle\lim_{(x,y)\to(a,b)}f(x,y)=f(a,b) \iff \left(\displaystyle\lim_{(x,y)\to(a,b)}g(x,y)=0\right)\land \left(\displaystyle\lim_{x\to a}f_{b}(x)=f_b{(a)}\right)$$ where $g:\mathbb{R}^2\to \mathbb{R}$ defined by $g(x,y)=f(x,y)-f(x,b)$.

My Proof

Necessity Part

Let us first prove that, $$\displaystyle\lim_{(x,y)\to(a,b)}f(x,y)=f(a,b) \implies \left(\displaystyle\lim_{(x,y)\to(a,b)}g(x,y)=0\right)\land \left(\displaystyle\lim_{x\to a}f_{b}(x)=f_b{(a)}\right)$$ Let $((x_n,y_n))_{n\ge1}$ be any sequence converging to $(a,b)$. Then since $\displaystyle\lim_{(x,y)\to(a,b)}f(x,y)=f(a,b)$, by definition we have, $$(\forall \varepsilon>0)(\exists n_1(\varepsilon)\in\mathbb{N})[n\ge n_1(\varepsilon)\implies \left\lvert f(x_n,y_n)-f(a,b)\right\rvert<\varepsilon]\tag{1}$$ Furthermore since $f_b$ is continuous at $x=a$ (I am skipping the proof of it for now) we have, $$(\forall \varepsilon>0)(\exists n_2(\varepsilon)\in\mathbb{N})[n\ge n_2(\varepsilon)\implies \left\lvert f(x_n,b)-f(a,b)\right\rvert<\varepsilon]\tag{2}$$ What remains is to prove that, $\displaystyle\lim_{(x,y)\to(a,b)}g(x,y)=0$. For this observe that from $(1)$ and $(2)$ we can write, $$\left(\forall \dfrac{\varepsilon}{2}>0\right)\left(\exists n_1\left(\dfrac{\varepsilon}{2}\right)\in\mathbb{N}\right)\left[n\ge n_1\left(\dfrac{\varepsilon}{2}\right)\implies \left\lvert f(x_n,y_n)-f(a,b)\right\rvert<\dfrac{\varepsilon}{2}\right]$$ $$\left(\forall \dfrac{\varepsilon}{2}>0\right)\left(\exists n_2\left(\dfrac{\varepsilon}{2}\right)\in\mathbb{N}\right)\left[n\ge n_2\left(\dfrac{\varepsilon}{2}\right)\implies \left\lvert f(x_n,b)-f(a,b)\right\rvert<\dfrac{\varepsilon}{2}\right]$$ Now taking $N\left(\dfrac{\varepsilon}{2}\right)=\max\left(n_1\left(\dfrac{\varepsilon}{2}\right),n_2\left(\dfrac{\varepsilon}{2}\right)\right)$ we get, $$\left(\forall \dfrac{\varepsilon}{2}>0\right)\left(\exists N\left(\dfrac{\varepsilon}{2}\right)\in\mathbb{N}\right)\left[n\ge N\left(\dfrac{\varepsilon}{2}\right)\implies \left\lvert f(x_n,y_n)-f(a,b)\right\rvert<\dfrac{\varepsilon}{2}\right]\tag{3}$$ $$\left(\forall \dfrac{\varepsilon}{2}>0\right)\left(\exists N\left(\dfrac{\varepsilon}{2}\right)\in\mathbb{N}\right)\left[n\ge N\left(\dfrac{\varepsilon}{2}\right)\implies \left\lvert f(x_n,b)-f(a,b)\right\rvert<\dfrac{\varepsilon}{2}\right]\tag{4}$$ From $(3)$ and $(4)$ and using Triangle Inequality we get, $$\left(\forall \varepsilon>0\right)\left(\exists N\left(\dfrac{\varepsilon}{2}\right)\in\mathbb{N}\right)\left[n\ge N\left(\dfrac{\varepsilon}{2}\right)\implies \left\lvert f(x_n,y_n)-f(x_n,b)\right\rvert<\varepsilon\right]$$ In other words, $$\left(\forall \varepsilon>0\right)\left(\exists N\left(\dfrac{\varepsilon}{2}\right)\in\mathbb{N}\right)\left[n\ge N\left(\dfrac{\varepsilon}{2}\right)\implies \left\lvert g(x_n,y_n)-g(a,b)\right\rvert<\varepsilon\right]$$ and this is what we wanted to prove.

Sufficiency Part

Now we try to prove that, $$\displaystyle\lim_{(x,y)\to(a,b)}f(x,y)=f(a,b) \impliedby \left(\displaystyle\lim_{(x,y)\to(a,b)}g(x,y)=0\right)\land \left(\displaystyle\lim_{x\to a}f_{b}(x)=f_b{(a)}\right)$$ Let $((x_n,y_n))_{n\ge1}$ be any sequence converging to $(a,b)$. Observe that since $g$ is continuous at $(a,b)$ we have, $$\left(\forall \varepsilon>0\right)\left(\exists n_1\left(\varepsilon\right)\in\mathbb{N}\right)\left[n\ge n_1\left(\varepsilon\right)\implies |g(x_n,y_n)-g(a,b)|<\varepsilon\right]$$ In other words, $$\left(\forall \varepsilon>0\right)\left(\exists n_1\left(\varepsilon\right)\in\mathbb{N}\right)\left[n\ge n_1\left(\varepsilon\right)\implies |f(x_n,y_n)-f(x_n,b)|<\varepsilon\right]\tag{5}$$ Also, since $f_b$ is continuous at $a$, we have, $$(\forall \varepsilon>0)(\exists n_2(\varepsilon)\in\mathbb{N})[n\ge n_2(\varepsilon)\implies |f(x_n,b)-f(a,b)|<\varepsilon]\tag{6}$$ Now observe that from $(5)$ and $(6)$ we can conclude respectively that, $$\left(\forall \dfrac{\varepsilon}{2}>0\right)\left(\exists n_1\left(\dfrac{\varepsilon}{2}\right)\in\mathbb{N}\right)\left[n\ge n_1\left(\dfrac{\varepsilon}{2}\right)\implies |f(x_n,y_n)-f(x_n,b)|<\dfrac{\varepsilon}{2}\right]$$ $$\left(\forall \dfrac{\varepsilon}{2}>0\right)\left(\exists n_2\left(\dfrac{\varepsilon}{2}\right)\in\mathbb{N}\right)\left[n\ge n_2\left(\dfrac{\varepsilon}{2}\right)\implies |f(x_n,b)-f(a,b)|<\dfrac{\varepsilon}{2}\right]$$
Now taking $N\left(\dfrac{\varepsilon}{2}\right)=\max\left(n_1\left(\dfrac{\varepsilon}{2}\right),n_2\left(\dfrac{\varepsilon}{2}\right)\right)$ from the above two statements (using Triangle Inequality) we can conclude that, $$\left(\forall\varepsilon>0\right)\left(\exists N\left(\dfrac{\varepsilon}{2}\right)\in\mathbb{N}\right)\left[n\ge N\left(\dfrac{\varepsilon}{2}\right)\implies |f(x_n,y_n)-f(a,b)|<\varepsilon\right]$$ Since $((x_n,y_n))_{n\ge1}$ was arbitrary, we are done.

Answer

Yes, your proof is entirely correct, allthough it is hard to read because of your quantor notation. But mathematically it is completely valid. Good Job!