Counterintuitive Charecterization of Openness in Euclidian Topology I came across an interesting problem and after playing with it for a few days I made some progress but resorted to looking up some solutions by others. My question is about their solutions, so please read along to the end before answering or commenting.
The original problem:

Prove or disprove: Consider $\mathbb{R}^2$ with the Euclidean topology
and let $X$ be a subset. If for every $a \in X$ and $v \in
 \mathbb{R}^2$ there exists $d > 0$ such that $a + tv \in X$ for every
$0 \leq t < d$ then $X$ is open.

If you haven't seen this problem before you might want to stop now and try it before I spoil the fun, since I am about to list some solutions below. It is a fun problem.
Why the problem is interesting:
The standard method of showing a set is open is to cover the set in open balls. The hypothesis of this problem naively makes us think we can do that. Given a point $a$ in $X$ and a point $v$ in the plane, by hypothesis there is some open ray starting at $a$ in the direction of $v$ that is contained in $X$. So if we center a circle around $a$ and construct an open ray in the direction of every point on the circle, we hope to have created something that can be reduced to an open disk around $a$ and contained in $X$. [Technically to use the hypothesis as stated we first need to  relocate the set $X$ so that $a$ is located at the origin, this makes $a+tv$ actually represent the ray from $a$ in the direction of $v$, but the ability to construct an open disk around $a$ doesn't depend on the location of $X$ in the plane, so I won't mention this detail again.]

This intuition motivated me to try and prove the claim: so let $a \in X$ and let $S$ be a circle of radius one centered on the point $a$. Let $v$ be an arbitrary point on $S$ and let $$R_v = \{ a+tv \, | \, 0 \leq t < d_v \}$$ be the set of points on the open ray in the direction of $v$ that exists by hypothesis. Now let $$D = \{d_v \, | \, v \in S \}.$$ If $d = \mathrm{min}(D)$ then we can consider $$R_{v_d} =\{ a+tv \, | \, 0 \leq t < d \}$$ and finally $$\bigcup_{v \in S} R_{v_d}$$ is an open disk centered on $a$ contained in $X$. Problem: most members of this community will have already noticed the issue, the existence of $\mathrm{min}(D)$ is not guarunteed, and it is fairly clear we will not be able to establish such an existence. At this point I abandon trying to prove the claim and begin looking for a set which fails to be open, satisfies the hypothesis, and where $\mathrm{min}(D)$ does not exist.
The question: I found hunting for a counterexample much more difficult than attempting to prove the proposition. I have built a large tool belt of standard methods and techniques for trying to prove openness, and trying to prove claims in general, but trying to find a counterexample seemed much more creatively demanding. I am looking for a set that is not open, that satisfies the hypothesis, and something where $\mathrm{min}(D)$ will not exist, since if $\mathrm{min}(D)$ always existed the theorem would be true. After struggling for a few days I looked up solutions: these are what I found:
Solution 1: $\mathbb{R}^2 - \{(x,y) \, | \, x^2 + y^2 = 1 \wedge (x,y) \neq (0,1) \}$. In words: the plane kickout a circle and add back in a point.
Solution 2: $\mathbb{R}^2 - \{(x,y) \, | \, y = x^2 \wedge x > 0 \}$
In both solutions there is a point that causes the set to not be open, (0,1) and (0,0) respectively so I agree neither set is open. I intuitively see how the constructions are intended to meet the hypothesis yet fail to have a minimum $d$: as we swing the ray around, the shape of the curve removed forces the ray to be smaller and smaller, hopefully causing no minimum. I think any counter example to this problem will involve a point with a removed arc leading up to it contained in some larger open set (perhaps $\mathbb{R}^2$ itself). But what I am having trouble fully convincing myself of is:

Is there not one single small direction when we pass the ray from
below the curve removed to above the curve removed where the length of
the ray must be 0? How can we prove that these solutions fully satisfy
the hypothesis: that is, for every direction we have the open ray of
non zero length?

What have a tried:
Without fail someone will ask it - so lets just cover it now. I worked with solution 1 and considered the point (0,1). I can parameterize the ray from the point to the removed circle as $$r(t) = (0,1) + t*(v - (0,1))$$ where $v = (cos(x), sin(x)), 0 \leq x < 2\pi$.
It is sufficient to derive an expression for the length of $r(t)$ and show that I can swing $r(t)$ around a full $2\pi$ radians and always have nonzero length.
I am not convinced that it can move beyond the point of being parallel to the x axis without going to 0!
 A: 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 $(0,d)$.
