Proving discontinuity of a map between $\mathbb{R}^2$ and $\mathbb{R}$, at a single point I am trying to show that, no matter what $c \in \mathbb{R}$ we take in the below, $f$ is not continuous at $0$:
$$f(x)=\left\{\begin{array}{ll}
\frac{x_1^2-x_2^2}{x_1^2+x_2^2} & \text { if } x \neq 0 \\
c & \text { if } x=0
\end{array}\right.
$$
Is my attempt correct?
Consider first some arbitrary $c\in \mathbb{R} \backslash \{-1, 1\}$.  Set $\epsilon:= \min\{|c-1|, |c+1|\}$. Consider some arbitrary $\delta>0$. Choose $x:= (x_1, x_2)$ such that $d(x, 0)=\sqrt{x_1^2+x_2^2}<\delta$.  Now, note that: $|f(0)-f(x)|=|c-\frac{x_1^2-x_2^2}{x_1^2+x_2^2}|=|\frac{(x_1^2+x_2^2)c-x_1^2+x_2^2}{x_1^2+x_2^2}|=|\frac{x_1^2c+x_2^2c+x_2^2-x_1^2}{x_1^2+x_2^2}|=|\frac{x_1^2(c-1)+x_2^2(c+1)}{x_1^2+x_2^2}|\geq\frac{x_1^2|(c-1)|+x_2^2|(c+1)|}{|x_1^2+x_2^2|} > \frac{(x_1^2+x_2^2)\min\{|c-1|,  |c+1|\}}{|x_1^2+x_2^2|}=$ $\min \{|c-1|, |c+1|\}$.
Next, consider $c=1$. Set $\epsilon:=1$. Consider some arbitrary $\delta>0$. Choose $x:= (x_1, x_2)$ such that $x_1=0$, and $x_2=\sqrt\frac{\delta}{2}$. Then $d(x, 0)<\delta$, but $|f(0)-f(x)|=|c-\frac{x_1^2-x_2^2}{x_1^2+x_2^2}|=|\frac{2x_2^2}{x_1^2+x_2^2}|={2}>1.$
Finally, consider $c=-1$. Set $\epsilon:=1$. Consider some arbitrary $\delta>0$. Choose $x:= (x_1, x_2)$ such that $x_2=0$, and $x_1=\sqrt\frac{\delta}{2}$. Then $d(x, 0)<\delta$, but $|f(0)-f(x)|=|c-\frac{x_1^2-x_2^2}{x_1^2+x_2^2}|=|\frac{2x_1^2}{x_1^2+x_2^2}|={2}>1.$

The problem with your approach lies in the fact that if for real numbers we have $a>b$ then we cannot deduce that $|a|>|b|$. In your case, while it's true that $x_1^2(c-1)+x_2^2(c+1) \geq (x_1^2+x_2^2)\cdot \min\{c-1,c+1\}$, you cannot take absolute values of both sides because you have no guarantee that they are both positive. As an example, for $c=-1$ the inequality gives $-2x_1^2 \geq -2(x_1^2+x_2^2)$ but when you take absolute values you get $x_1^2\geq x_1^2+x_2^2$ which is obviously not correct.
The simplest way to attack this problem is to find two paths that give rise to two different limits which will prove that the limit of $f$ at $(0,0)$ does not exist.
