Intro: The following two questions are from my exam preparation sheet, it is not mandatory and will not be accredited (or improve marks and the like). There won't be a correction, merely an online solution to it. Since we had very few exercises related to the verification of open and/or closed sets, I want to ask here if my thinking and argumentation is correct. Pardon me if it happens to be a yes/no comment type of question.
Similar: I have read Simple question about closed sets and found it very helpful already.
Problem: Let $f,g: \mathbb{R}^n \to \mathbb{R}$ be two continuous functions. Show that the set $$X_1= \lbrace x \in \mathbb{R}^n \mid f(x)=g(x+(1, \dots , 1))^2 \rbrace \text{ is closed } \\ X_2= \lbrace x \in \mathbb{R} \mid f(x) > 1-g(x^2) \rbrace \text{ is open}$$
My Approach: For $X_1$ let $x_n$ be a sequence in $X_1$ such that $\lim_{n \to \infty}x_n = x' \in \mathbb{R}^n$ then I have $$\lim_{n \to \infty} f(x_n)=\lim_{n \to \infty} g(x_n+(1, \dots, 1))^2 \overset{1)}{=}g(x'+(1 , \dots , 1))^2=f(x') \in X_1 $$
Is this the right approach to show that the set is closed?
1) $g$ is continuous
For $X_2$ my initial thought was to come up with a contradiction, defining a convergent sequence such that $\lim_{n \to \infty} x_n =0 \in \mathbb{R}$ which would have left to the inequality $0 > 1$ but I believe this is the wrong approach. So inspired by the linked question above I came up with: $$f(x)>1-g(x^2) \iff h(x):=f(x)+g(x^2)-1>0 \implies h(x) \in (0,\infty)$$ the defined function $h(x)$ is continuous because it's a composition of continuous functions and is an element of the open set $(0, \infty)$
Providing this answer so that the question, which was answered by the OP and in the comments, can be closed.