I'm seriously struggling with understanding how to do these problems.
I start with the theorem:
$\forall\,\epsilon >0,\ \exists\,\delta>0$ s.t.
$d[(x,y),(c,d)] < \delta$ $\implies$ $d(f(x,y), f(c,d)) < \epsilon$
I know the above is just the general theorem but I think it's good to have visible.
Thank you in advance for your help.
The so-called "theorem" is indeed a definition which goes as follows: $f:\mathbb{R}^2 \to \mathbb{R}$ is continuous in $(a,b)\in\mathbb{R}^2$ if there for any $\epsilon>0$ exists a $\delta>0$ such that for any $(x,y)\in\mathbb{R}^2$ with $d((x,y),(a,b))<\delta$ then $d(f(x,y),f(a,b))<\epsilon$.
Now to show that $f(x,y)=\sqrt{x^2+y^2}$ is indeed continuous everywhere on $\mathbb{R}^2$, we first fix an arbitrary $(a,b)$ and $\epsilon>0$. Note that $f$ is indeed the Euclidean norm on $\mathbb{R}^2$, so we can use the reverse triangle inequality. For any $(x,y)\in\mathbb{R}^2$ with $|(x,y)-(a,b)|<\delta=\epsilon$ \begin{align} d(f(x,y),f(a,b))&=|f(x,y)-f(a,b)| \\ &= |\sqrt{x^2+y^2}-\sqrt{a^2+b^2}|\\ &= |\, |(x,y)| - |(a,b)|\, | \\ &\leq |(x,y)-(a,b)| \\ &< \epsilon, \end{align} proving that $f$ is continuous in $(a,b)$ and since this was arbitrarily chosen we conclude that $f$ is everywhere continuous. This obviously assumes that both domain and co-domain of $f$ are equipped with the corresponding Euclidean metrics.