Proof explanation to see that subset of $\mathbb{R}^2$ is not path connected. Consider the following example from http://www.mathcounterexamples.net/a-connected-not-locally-connected-space/ .
The part that I don't understand it's the bold text. What is $\arg⁡(B)$?
And why $\arg⁡(1,1)$ is an isolated point of $\arg⁡(B)$?
I greatly appreciate any assistance you may provide.

$$A = \bigcup_{n \ge 1} [(0,0),(1,\frac{1}{n})] \text{ and } B = A \cup (\frac{1}{2},1]$$
For the proof that $B$ is not path connected, suppose that $\gamma$ is a path joining the point $(3/4,0)$
  to the point $(1,1).$ We denote $\arg \gamma(t)$ the angle of $\gamma (t)$ with the $x$-axis. $\arg \gamma(t)$ is a continuous function of $t \in [0,1].$ Hence it range is connected.
This is in contradiction with the fact that $\arg⁡(1,1)$ is an isolated point of $\arg⁡(B).$

 A: Short answer

source code of the figure
$$\arg(B) = \left\{ \mbox{set of all possible angles of oblique lines} \left[ (0,0),\left( 1,\frac{1}{n} \right) \right] \mid n \in \Bbb{N} \right\}$$
One easily observes from the above figure that the angle $\dfrac\pi4$ is an isolated point of $\arg(B)$.
Detailed Explanation
Note that $$\arg(B) = \bigcup_{n \ge 1} \left\{ \arctan\frac1n \right\} \cup \{0\}.$$
Observe that $\arctan$ is a continuous function, so that $$\arctan\frac1n \xrightarrow[n \to \infty]{} \arctan 0 = 0.$$
The first oblique line $[(0,0),(1,1)]$ and the set difference between $B$ and this oblique line
\begin{align}
&\quad |\arg(1,1) - \arg(B \setminus [(0,0),(1,1)])| \\
&\ge \arctan 1 - \arctan\frac12\\
&= \frac\pi4 - \arctan\frac12 \\
&\approx 0.322 \tag{cor. to 3 sig. fig.}\end{align}
A: The proof needs a little tweak to avoid the issue of $\operatorname{arg}$
not being defined on all of $B$.
Let $I= ({1 \over 2} , 1] \times \{0\}$.
Let $t_0 = \sup \{ t | \gamma(t) \in I \}$. We have $[\gamma(t_0)]_1 \ge {1 \over 2}$ and hence there is some $\delta>0$ such that $[\gamma(t)]_1 \ge {1 \over 4}$
for $t \in [0,t_0+ \delta]$. Also, $\gamma(t) \in A$ for $t > t_0$ and
$\gamma(t_0) \in \overline{I}$.
For $ x\in [{1 \over 4} , \infty) \times \mathbb{R}$ define
$\alpha(x) = {x_2 \over x_1}$ (a proxy for $\operatorname{arg}$). Note that
$\alpha$ is continuous.
Note that $\alpha(\overline{I}) = \{0\}, \alpha(A \cap [{1 \over 4} , \infty)) = \{ {1 \over n}\}_n$.
Hence $\alpha \circ \gamma ([t_0,t_0+\delta]) \subset \{0\} \cup \{{1\over n}\}_n$. However, since $\alpha \circ \gamma$ is continuous, the set
$\alpha \circ \gamma ([t_0,t_0+\delta]) $ must be an interval containing
the set $[0, \alpha(\gamma(t_0+\delta))]$ which is a contradiction
since $\{0\} \cup \{{1\over n}\}_n$ contains no non trivial intervals.
