I didn't check your proof, but here is a much easier way to prove your result.
Let $$L_1:=\lim_{x\to a}f(x)\quad\text{and}\quad L_2=\lim_{x\to a}g(x).$$
Let $(x_n)$ be a sequence that converges to $a$. For all $n$ we have, $$ f(x_n)\leq g(x_n),$$ and thus $$L_1=\lim_{n\to\infty }f(x_n)\leq \lim_{n\to\infty }g(x_n)=L_2,$$
which proves the claim.
Here is my own proof. Please tell me what if any steps of this seem "imprecise" to you.
If $f$ has an "epsilon-delta limit" of $l$ at $x_0$, then $f$ has a "sequence limit" of $l$ at $x_0$: Let $(x_n)$ be a sequence in $D_f\setminus\{x_0\}$ such that $\displaystyle\lim_{n \to \infty} x_n=x_0$. We need to show that $\displaystyle\lim_{n \to \infty} f(x_n)=l$. Let $\epsilon>0$. Then, since $f$ has an "epsilon-delta limit" of $l$ at $x_0$, there exists $\delta>0$ such that for all $x \in D_f$ satisfying $0 <|x-x_0|<\delta$, we have $|f(x)-l|<\epsilon$. Since $\displaystyle\lim_{n \to \infty}x_n=x_0$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $|x_n-x_0|<\delta$. Then $0<|x_n-x_0|<\delta$ for all $n \geq N$, so $|f(x_n)-\ell|<\epsilon$ for all $n \geq N$. We have shown that given $\epsilon>0$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, $|f(x_n)-l|<\epsilon$. Thus $\displaystyle\lim_{n \to \infty}f(x_n)=l$.
If $f$ has a "sequence limit" of $l$ at $x_0$, then $f$ has an "epsilon-delta limit" of $l$ at $x_0$: We will prove the contrapositive, i.e., if $f$ does not have an "epsilon-delta limit" of $l$ at $x_0$, then $f$ does not have a "sequence limit" of $l$ at $x_0$. If $f$ does not have an "epsilon-delta" limit of $l$ at $x_0$, then there exists $\epsilon>0$ such that for all $\delta>0$, there exists $x \in D_f$ satisfying $0<|x-x_0|<\delta$, and yet $|f(x)-l| \geq \epsilon$. To show that $f$ does not have a "sequence limit" of $l$ at $x_0$, we must exhibit a sequence $(x_n)\in D_f \setminus\{x_0\}$ such that $\displaystyle\lim_{n \to \infty}x_n=x_0$ and yet $\displaystyle\lim_{n \to \infty}f(x_n) \neq l$.
We construct the sequence as follows. Setting $\delta_1=1$, there exists $x_1\in D_f$ with $0<|x_1-x_0|<\delta_1=1$ such that $|f(x_1)-l| \geq \epsilon$. (Note then that $x_1 \neq x_0$, so in fact $x_1 \in D_f \setminus\{x_0\}$.) Similarly, setting $\delta_2=\frac{1}{2}$, there exists $x_2 \in D_f$ with $0<|x_2-x_0|<\delta_2=\frac{1}{2}$ such that $|f(x_2)-l| \geq \epsilon$. (Once again, $x_2 \neq x_0$, so $x_2 \in D_f \setminus\{x_0\}$.) Now suppose we have chosen $x_1,x_2,\dots,x_n \in D_f \setminus\{x_0\}$ such that $|x_i-x_0|<\frac{1}{i}$ and $|f(x_i)-l| \geq \epsilon$ for all $1 \leq i \leq n$. Setting $\delta_{n+1}=\frac{1}{n+1}$, there exists $x_{n+1} \in D_f$ with $0 \leq |x_{n+1}-x_0|<\frac{1}{n+1}$ such that $|f(x_{n+1})-l| \geq \epsilon$, so we let $x_{n+1}$ be the next term of the sequence. By the axiom of dependent choice, we can define a sequence $(x_n) \in D_f \setminus\{x_0\}$ such that $0<|x_i-x_0|<\frac{1}{i}$ but $|f(x_i)-l| \geq \epsilon$ for all $i \geq 1$.
Claim: $\displaystyle\lim_{n \to \infty} x_n=x_0$. Let $\eta>0$. Then there exists a natural number $N>\frac{1}{\eta}$. For all $n \geq N$, $|x_n-x_0|<\frac{1}{n}\leq\frac{1}{N}< \frac{1}{1/\eta}=\eta$. So $\displaystyle\lim_{n \to \infty} x_n=x_0$.
Claim: $\displaystyle\lim_{n \to \infty} f(x_n)=x_0$. Let $\epsilon$ be as above. Choose any $M \in \mathbb{N}$. Then for any $m \geq M$, $|f(x_m)-l| \geq \epsilon$. (Note: this is stronger than we need. We only need to show there exists $m \geq M$ satisfying that equation.) Hence $\displaystyle\lim_{n \to \infty} f(x_n) \neq l$.
It is good to be careful when one is first learning to write proofs, but as we can see above, sometimes including all the details makes proofs long and tedious. For instance, if I were writing this proof elsewhere, I would said that it is clear that the sequence $(x_n)$ satisfying $|x_i-x_0|<\frac{1}{i}$ for all $i$ converges to $x_0$. I would have also said it's clear that $\displaystyle\lim_{n \to \infty} f(x_n) \neq l$ is clear if each term $f(x_i)$ satisfies $|f(x_i)-l| \geq \epsilon$. And when defining that sequence $(x_n)$ inductively, I might not have written out the whole induction argument, but rather said "and so on..." These sorts of shortcuts are not necessarily imprecise. They are just an indicator that the fact that is claimed is routine to verify, and so the author has not taken the time and space to spell out all the details.
Best Answer
The limit you have defined is perfectly valid. And what you are trying to show in such a case is that for every $y$ $f(x,y)$ converges to some limit as $x \to c$. Since you are very likely to obtain different limit for each $y$, the resulting collection of limits is naturally forms a function $g(y)$.
The next question to ask is how does the convergence run for each $y$? If the same $\delta$ can work for a given $\varepsilon$ and all $y$, then the convergence is uniform. Expressed formally, convergence to $g(y)$ is uniform if for every $\varepsilon >0$ there exists $\delta > 0$ (depending on $\varepsilon$ but not on $y$) such that, $$ | f(x,y)-g(y) |< \varepsilon \text{ for all } y \text{ whenever } |x-c| < \delta.$$
Uniform convergence is helpful because you can start deriving properties of $g$ based on properties of $f(x,y)$. Of course this can also happen when convergence is not uniform, but that often requires more work. In particular if $f(x,y)$ is continuous in $y$ and convergence is uniform then $g(y)$ is also continuous.
Convergence of $f(x,y) \to 3y$ as $x \to 3$ is not uniform when $y$ can take any real value because for large $y$, $x$ has to get closer to $3$ to achieve the same level of approximation. But if you restrict your interest to $y$ in a bounded interval (e.g. $-100 \leqslant y \leqslant 100$) then the worst case arises for $y=\pm 100$ and so convergence is uniform.