Let S = $ \{(x,y) \in \mathbb{R}^{2}| x^{2} + y^{2} =1 \} $
I want to prove that this set is closed or by showing that it's complement is open. Now my proof becomes show that $S^{c}$ = $ \{(x,y) \in \mathbb{R}^{2}| x^{2} + y^{2} \neq 1 \} $ is open.
I want to prove this by finding a continuous function $f$ given by $f(x,y) = x^{2} + y^{2}$ where S = $f^{-1}$ $(-\infty, -1) \cup (-1,1) \cup (1, \infty)$
Since $(-\infty, -1) \cup (-1,1) \cup (1, \infty)$ are open and $f$ is continuous, $S^{c}$ is open and therefore set S is closed.
Is my reasoning correct. Please correct me if I am wrong in any part.
Your reasoning is correct and you've got a correct justification. However, I don't quite see why you are afraid of that $-1$. Indeed, $f\ge 0$. So you can use $$\mathbb R \setminus\{1\}, \ \ (-1,1)\cup (1, \infty)$$
or even $$(-n, 1)\cup (1, \infty) $$
for any $n<0$. The preimage of this map will be $S^c$.
Or you can use the fact that for a continuous function, preimage of a closed set is a closed set. Thus $S = f^{-1}(\{1\})$ is closed.