if there is a function which is defined in some open neighborhood of the closed unit disk, analytic in the open unit disk, complex differentiable at $z=1$, but not differentiable at each of a sequence of real $x_n>1$ which converge to $1$.
Yes. Let $S$ be the set $\{1+re^{it}: r\ge 0, |t|\le \pi/3\}$ (angle/sector/cone or what you call it). In the domain $\mathbb C\setminus S$ the function $(1-z)^{3/2}$ has a single valued branch with $(1-0)^{3/2}=1$. On the boundary of $S$ this branch takes real, nonpositive values. Let
$$f(z)=\begin{cases} (1-z)^{3/2} \quad & z\notin S \\
-|1-z|^{3/2} & z\in S
\end{cases}$$
The function $f$ is continuous on $\mathbb C$, analytic in the open unit disk, complex differentiable at $1$ (with $f'(1)=0$), but is not holomorphic in any neighborhood of $1$, since it is real-valued in $S$.
The point $1$ is not a limit point of the set, because there is a neighbourhood of $1$ such that the only point in the set in that neighbourhood is $1$. Use, for example, the interval $(0.9,1.1)$.
In fact no point in the set is a limit point of the set. Around the point $\frac{1}{n}$, you can put the neighbourhood $(\frac{1}{n}-\frac{1}{4^n},\frac{1}{n}+\frac{1}{4^n})$ that contains no point of our set other than $\frac{1}{n}$.
To show that $0$ is a limit point of our set, take an open interval $(-\epsilon,\epsilon)$ about $0$, where $\epsilon$ is small (possibly very small) positive. There is an integer $N$ such that $\frac{1}{N}\lt \epsilon$. Then every point $\frac{1}{n}$ in our set with $n\ge N$ is in that neighbourhood.
So every open neighbourhood of $0$ contains a point of our set, indeed infinitely many points of our set.
Much more informally, we can get arbitrarily close to $0$ from within our set.
A limit point of a set may or may not belong to the set. For example, let $S=(0,1)$, that is, all real numbers $x$ such that $0\lt x\lt 1$. The number $1$ is a limit point of $S$, and is not in $S$.
Let $T=[0,1]$. The point $1$ is a limit point of $T$ and is in $T$.
Best Answer
One kind of neighborhood is an open disk around $z_0$. In your example, $z_0$ is $0.1$ from the line $Re(z)=1$, so any disk of radius less than $0.1$ will work. It doesn't depend on the imaginary part ($y$) because the division line doesn't depend on $y$.
If the set were the points for which $2Re(z) \gt Im(z)$, it is again an open set. Your $z_0=1.1+2i$ is in this set too. But the distance from the dividing line depends on $y$ in this case because the line is sloping.