Background
Let $\{B_t\}_{t\ge 0}$ be Brownian motion in $\mathbb R^d$. For $D\subset\mathbb R^d$, let $\tau_D=\inf\{t>0:B_t\in D\}$ be the hitting time of $D$ for $B_t$. For $D\subset\mathbb R^d$ open, a point $x\in\partial D$ is said to be regular if $\mathbb P_x(\tau_{D^c}=0)=1$; that is, if when the Brownian motion is started from $x$ it hits $D^c$ instantaneously (possibly by returning to $x$). An open set $D$ is said to be regular if all of its boundary points are.
Motivation
If $d=1$, every set is regular because for any $\varepsilon>0$ the Brownian motion $B_t$ will return to its starting point at infinitely many $t$ in the interval $(0,\varepsilon)$.
If $d\ge 2$, I believe the unit ball $\mathbb B=\{x\in\mathbb R^d:|x|<1\}$ is regular and simply connected, the annulus $\mathbb A=\{x\in\mathbb R^d:1<|x|<2\}$ is regular but not simply connected, and the punctured ball $\mathbb B\setminus\{0\}$ is neither.
If $d\ge 3$, then I believe $\mathbb B\setminus\left([0,1)\times\{0\}^{d-1}\right)$ should not be regular, because e.g. the second two coordinates of the Brownian motion will both be zero concurrently with probability zero.
However, when $d=2$ I haven't been able to think of an open set which is simply connected but not regular. If we look at $\mathbb B\setminus\left([0,1)\times\{0\}\right)$ again, this time I don't think it works. In response to the comments, we'll look at what's going on here more carefully. Let $B_t=(B^1_t,B^2_t)$.
For boundary points $(a,0)$ with $0<a<1$, as $\varepsilon\to0$ the probability that $B_t^1\in(0,1)$ for all times $t\in(0,\varepsilon)$ approaches one by continuity of probability and continuity of Brownian motion. And conditional on the event $\{\forall t\in(0,\varepsilon), B_t^1\in(0,1)\}$ the probability that $B_t$ hits the boundary at some point in $(0,\varepsilon)$ is one, since $B_t^2$ is zero for infinitely many $t\in(0,\varepsilon)$ with probability one.
So the only possible irregular boundary point of $\mathbb B\setminus\left([0,1)\times\{0\}\right)$ is $(0,0)$. Here I don't have a proof of regularity, just a heuristic. Namely, $B_t^2$ will be zero at infinitely many times on the interval $(0,\varepsilon)$, and because $B^1$ and $B^2$ are independent I expect that "half" of those times the first coordinate will be non-negative.
Question
Is every simply connected open subset of $\mathbb R^2$ regular, or is there an counterexample that I'm not thinking of?
The regularity of every boundary point of an open set in $\mathbb R^2$ is in fact strongly related to connectedness. However, that's more a property of the complement of the set:
Problem 4.2.16 in [1] which is:
Let $D\subset\mathbb R^2$ be open, and suppose that $a\in\partial D$ has the property that there exists a point $b\not=a$ in $\mathbb R^2\setminus D\,,$ and a simple arc in $\mathbb R^2\setminus D$ connecting $a$ to $b\,.$ Show that $a$ is regular.
The solution is provided in [1] section 4.5.
The unit disc minus the line segment $[0,1)\times\{0\}$ clearly satisfies the properties of $D$ in that problem.
[1] I. Karatzas, S. Shreve, Brownian Motion and Stochastic Calculus.