Counterintuitive Charecterization of Openness in Euclidian Topology

This is just simple algebra when you write down what the statement means very explicitly in coordinates. Let me consider your second example. So we want to show that for any $v=(a,b)\in\mathbb{R}^2$, there exists $d>0$ such that for all $t\in [0,d)$, $tv$ is not of the form $(x,x^2)$ for any $x>0$. Well, if $tv$ were of this form, that would mean that $ta>0$ and $tb=(ta)^2$. But we can solve the latter equation for $t$: assuming $t$ and $a$ are nonzero (which we can since we also need $ta>0$), it is only true if $t=b/a^2$. So, we can just let $d$ be any positive real number such that $b/a^2\not\in (0,d)$ (explicitly, if $b/a^2$ is positive you can take $d=b/a^2$ and otherwise $d$ can be anything).

The first example works similarly, except that instead of $tb=(ta)^2$ the equation is $(ta)^2+(1+tb)^2=1$ which you can again solve for $t$ to find only finitely many solutions for any fixed $(a,b)$ (except in the trivial case where $(a,b)=(0,0)$). So, you can choose $d$ to be small enough so that none of the solutions are in $(0,d)$.