Tell which of the following sets are closed/open/both/neither. prove your answer
S = $\{\textbf{x} \in \mathbb{R}^2: x_2>0 \}$
S = $\{\textbf{x} \in \mathbb{R}^2 : x_2> x_1\}$
S = $\{\textbf{x} \in \mathbb{R}^2 : x_1x_2 \neq 0\}$
S = $\{\textbf{x} \in \mathbb{R}^2 : x_1 = x_2 \}$
S = $\{\textbf{x} \in \mathbb{R}^2 : 0 < ||\textbf{x}|| <1 \}$
S = $\{\textbf{x} \in \mathbb{R}^2: x_2>0 \}$ is open because given arbitrary $\textbf{x} \in S$ we can fix $\delta = \min\{|x_1| , x_2\}$ it immediately implies that $B(\textbf{x}, \delta) \subset S$ and is not closed because consider sequence of vectors $\{[0,1/n]\}$ which is completely in $S$ but it converges to $[0,0]\not \in S$
S = $\{\textbf{x} \in \mathbb{R}^2 : x_2> x_1\}$ is open because given arbitrary $\textbf{x} \in S$ we can get distance from $x=y$ which is $\delta = \frac{|x_1-x_2|}{\sqrt{2}}$ which implies that $B(\textbf{x} , \delta) \subset S$ and is not closed because consider sequence of vectors $\{[1/n,0]\}$ which is completely in $S$ but it converges to $[0,0] \not \in S$
S = $\{\textbf{x} \in \mathbb{R}^2 : x_1 x_2 \neq 0\}$ is open because given arbitrary $\textbf{x} \in S$ we can set $\delta = \min \{|x_1| , |x_2|\}$ which implies that $B(\textbf{x} , \delta) \subset S$ and is not closed because sequence of vectors $\{[1/n,1]\}$ which is completely in $S$ but it converges to $[0,1] \in S$
S = $\{\textbf{x} \in \mathbb{R}^2 : x_1 = x_2 \}$ is not open because given any point s.t $x_1 = x_2$ there is no $\delta$ s.t $B(\textbf{x} , \delta) \subset S$ and since we have shown that $\mathbb{R}^2 - S = \{\textbf{x} \in \mathbb{R}^2 : x_2> x_1\} \cup \{\textbf{x} \in \mathbb{R}^2 : x_1> x_2\} $ now we have already show that first set is open in $2.$ and second set is analogous.so we have that $\mathbb{R}^2-S$ is open which implies that $S$ is closed
S = $\{\textbf{x} \in \mathbb{R}^2 : 0 < ||\textbf{x}|| <1 \}$ is open because given any $\textbf{x} \in S$ set $\delta = \min \{||\textbf{x}|| , 1- ||\textbf{x}||\}$ which implies that $B(\textbf{x} , \delta) \subset S$ and is not closed because consider sequence of vectors $\{[\frac{1}{n} , 0]\}_{n \geq 2}$ which is completely in $S$ but it converges to $[0,0] \not \in S$
You should back up your claims about that some $\delta$ works some more. You probably made some picture that convinces you, but that's not what a proof is.
E.g. for $1$: Let $\mathbf{x} \in S$ so that $x_2 >0$. Then define $\delta=x_2$ ($x_1$ is irrelevant in this case) and note that if $\|\mathbf{x'} - \mathbf{x}\| < \delta$, then $|x'_2 - x_2| \le \|\mathbf{x'} - \mathbf{x}\| < \delta$ so that $x'_2 > 0$ (if $x'_2 \le 0$, $|x'_2 - x_2| = x_2 + (-x'_2) \ge x_2=\delta$ which is not the case), so $\mathbf{x}' \in S$, and $B(\mathbf{x},\delta) \subseteq S$ as claimed.
You do need to some computations and estimations if you're working from the definitions as you're probably assigend to do. Later you'll probably develop more theory and then the openness and closedness of these sets will be easier to prove, at least less tedious. E.g. the projections are continuous so $S=\pi_2^{-1}[(0,\infty)]$ is open, for $1$ again. Done. And then if you've learned about connectedness you can say : in $\Bbb R^2$ this means that $S$ cannot be closed, but in the definition phase you can refute it by a sequence $\mathbf{x}_n:=(0,\frac{1}{n})$ as you do, assuming you prove that this sequence has the limit $(0,0)$ e.g. based on the observation that $\|\mathbf{x}_n\|=\frac{1}{n}$ and $\frac{1}{n} \to 0 \,(n \to \infty)$ etc. It'll depend on what you've already shown and covered about the properties of $\Bbb R^2$ and $\Bbb R$ so far.
Similar remarks can be made about the other claims. The end statements about closed and open are correct but make more work of the claims and their actual proofs (!), don't just claim that it holds, but base it on hard fact..