Consider the circular function $g:\mathbb{R}^{2} \to \mathbb{R}^{+}$, $g(x,y)=x^{2}+y^{2}$. Show that it is surjective and continuous.
Note
This post has been amended in accordance with the suggestions given below.
Surjectivity
Let $r \in \mathbb{R}^{+}$ be arbitrary. We want to find a $(x_{0},y_{0}) \in \mathbb{R}^{2}$ such that $g(x_{0},y_{0})=r$. Consider $x_{0}=\sqrt{r},y_{0}=0$ both of which are in $\mathbb{R}$. Then, \begin{align*} g(\sqrt{r},0)&=\left(\sqrt{r}\right)^{2}+0^{2} \\ &=r \end{align*} Since $r \in \mathbb{R}^{+}$ was arbitrary, it follows that $g:\mathbb{R}^{2} \to \mathbb{R}^{+}$, $g(x,y)=x^{2}+y^{2}$ is surjective.
Continuity. Let $\epsilon >0$. Let $x_{0},y_{0} \in \mathbb{R}^{2}$ be arbitrary. Then $|x^{2}-x_{0}^{2}|< \frac{\epsilon}{2}$ when $|x-x_{0}| < \delta$ and $|y^{2}-y_{0}^{2}| < \frac{\epsilon}{2}$ when $|y-y_{0}| < \delta$. Thus we have: \begin{align*} |g(x,y)-g(x_{0},y_{0})| &=|x^{2}+y^{2}-(x_{0}^{2}+y_{0}^{2})| \\ &=|(x^{2}-x_{0}^{2})+(y^{2}-y_{0}^{2})| \\ &\leq |x^{2}-x_{0}^{2}| +|y^{2}-y_{0}^{2}| \\ &\leq \frac{\epsilon}{2}+\frac{\epsilon}{2} \\ &=\epsilon \end{align*} as $(x,y) \to (x_{0},y_{0})$. Therefore, $g$ is continuous.
Your method suffices, but is a bit overly elaborate. An easier method is as follows: