The set $H=\{(x,y)\in \Bbb{R^2}:\;3x+2y=5 \},$ is a closed subset of $\Bbb{R^2}$ Kindly check if my proof is correct. Thanks for your time and efforts.
MY PROOF
For all $n\in \Bbb{N},$ let $(x_n,y_n)\in H$ such that $(x_n,y_n)\to (x,y),$ as $n\to \infty.$ We show that $(x,y)\in H.$
However, $(x_n,y_n)\in H$ implies $3x_n+2y_n=5,$ for all $n\in \Bbb{N}.$ As $n\to \infty, $
$$ 3x+2y=\lim\limits_{n\to\infty}(3x_n+2y_n)=\lim\limits_{n\to\infty}5=5 .     $$
Hence, $(x,y)\in H$ which implies that the set, $H=\{(x,y)\in \Bbb{R^2}:\;3x+2y=5 \},$ is a closed subset of $\Bbb{R^2}$
 A: Looks Good! Alternatively, take $\alpha=(x,y) \in \Bbb R^2 \setminus H$. Take $r=d(\alpha,H)>0$, the distance from the point $\alpha$ to the line $H$.  Then $$B(\alpha,r) \subset \Bbb R^2 \setminus H$$ showing $\Bbb R^2 \setminus H$ is open.
A: Maybe I'm overcomplicating this, but here's where I think your proof fails
Imagine that $H = \{(x, y)\in \Bbb{R}^2: x^2 + y^2 < 1\}$. And now I am going to apply your same argument

For all $n\in \Bbb{N},$ let $(x_n,y_n)\in H$ such that $(x_n,y_n)\to (x,y),$ as $n\to \infty.$ We show that $(x,y)\in H.$
However, $(x_n,y_n)\in H$ implies $x_n^2+y_n^2<1,$ for all $n\in \Bbb{N}.$ As $n\to \infty, $
$$\lim\limits_{n\to\infty} x_n^2 + y_n^2 = x^2 + y^2 < 1$$
Hence, $(x,y)\in H$ which implies that the set, $H=\{(x,y)\in \Bbb{R^2}:\;x^2+y^2<1 \},$ is a closed subset of $\Bbb{R}^2$

Do you see the problem?
A: I would modify your proof to make one tacit step explicit in the display: If $(x_n,y_n)\to(x,y)$, then
$$3x+2y=3\lim_{n\to\infty}x_n+2\lim_{n\to\infty}y_n=\lim_{n\to\infty}(3x_n+2y_n)=\lim_{n\to\infty}5=5$$
A: I think you should have $3x+2y$ in the 3 line from the bottom.... 
