Conceptual difficulty with using open balls to show whether or not a set is open or closed Here's an example question with solution to illustrate where I'm having difficulty understanding:

Consider the set $H := \{(x_1, x_2) : 1 \le x_1 \le 3, 2 < x_2 < 4\}$ in $\mathbb{R^2}$. Determine if $H$ is a closed set. Determine if $H$ is an open set.
Solution: We see that $(1,3) \in H$. But for $r > 0$, $B_r \left(1, 3 \right)$ contains a point $(1-\frac{r}{2}, 3)$ that is not in $H$. Then $H$ is not an open set. Further, we have $(2, 2) \in H^c$. But for $0 < r < 1$, $B_r(2, 2)$ contains the point $\left( 2, 2 + \frac{r}{2} \right) \notin H^c$. Then H is not a closed set.

The conclusions themselves, following finding the points in the open balls, are fine for me. But I am struggling conceptually with how we seem to be trivially finding that $B_r(1,3)$ contains $(1-\frac{r}{2}, 3)$. I have been given the definition of an open ball in $\mathbb{R^2}$:

For $r > 0 $ and $(a, b) \in \mathbb{R^2}$, the open ball centered at $(a, b)$ of radius $r$ is given by
$$B_r((a,b)) := \{(x,y) \in \mathbb{R}^2 : \lVert (x,y) - (a,b) \rVert < r \}.$$

I can't figure out how to reconcile this definition with any calculation that would easily enable us to find the point $(1 - \frac{r}{2}, 3)$ without either requiring a priori knowledge about the particular problem or using trial-and-error with various candidates for $(x,y)$. How are we reliably finding these values for $(x,y)$?
The tl;dr is that I understand the conclusions drawn once we obtain these points within the open balls, but can't figure out how to reliably obtain these points. I want to stress that while the example here illustrates my conceptual difficulty, I'm more confused as to how to generalize this to different sets.
Embarrassingly, I get the sense here that there's something trivial I'm just not grasping. Can someone help me find what I'm missing?
 A: $H=[1,3]×(2,4)$ is neither open nor closed.  Intuitively, you have a square that is missing the boundary on two sides, and contains the boundary on the other two.
But, closed sets contain all their boundary points; whereas open sets contain none.
Try drawing a picture.  You sort of get a square with upper and lower edges missing.
In terms of open balls, it's pretty easy to see that with interior points you can always squeeze one in, just by taking the radius less than the distance to the boundary.  Of course, you can't do that on boundary points, the distance is zero.
A: Conceptually, the solution takes an approach similar to what one might do to show that the interval $H = [1,2) \subset \mathbb R$ is neither open nor closed in $\mathbb R$. I'll do this in some detail to build analogy.
Let's focus first on proving $H$ is not open: we can visually see $1$ is on the edge/boundary of $H$, and is in $H$. Precisely, we could say that for every $r > 0$, the interval $(1-r, 1+r)$ centered at our boundary point 1 contains many points outside of $H$, for example $1-r/2$, but also $1-(r/1000)$. 
For non-closedness, we focus on the other "boundary" point of $H$, which is 2; however, $2 \not\in H$, so it's a good candidate for a point that is so close to being in $H$ that there are literally no points between 2 and $H$. For every $r > 0$ the interval $(2-r, 2+r)$ centered at 2 contain many points of $H$. Therefore, $H$ is not closed. 
The balls $B_r((a, b))$ play the same role as the intervals $(c - r, c +r)$, and I used the same strategy of finding boundary points of $H$ that are actually in $H$, and points that are in the boundary of the complement of $H$. As mentioned by Aston in the comments (forgive me, I don't have a Cyrillic keyboard on my phone), these boundary issues are often caused by $\leq$ versus $<$. But they could also be caused by things like starting with the region $\{ (x, y) \,:\, 1 \leq x \leq 1, 2 \leq y \leq 3 \}$ and removing one point, say $(0.7, 2.5)$, to give a region $X$. Now the same trick of picking boundary points like $(1,2)$ or $(1.6, 3)$ will let us show that $X$ isn't an open set. And if we focus on the removed point, any ball centered there necessarily contains points from $X$, so $X$ is also not closed.
A: I don't see a reason why the textbook uses $\frac{r}{2}$. It can use $\frac{r}{n}$ ($n$ is finite), or $\frac{1}{n}$. This is because $R$ has the Archimedean property, meaning given $x$ and $y$ in R and assume $y >$ 0, then there exists a positive integer $n$ such that $xn > y$. I agree with @Chris, since this is pretty easy to visualize this situation, especially with the standard metric, you should try drawing a picture.
About generalizing finding points to different sets: If you have a set and you define two different metrics on it, the open balls in each space will look different. If you can visualize the ball, you might be able to find the points intuitively, meaning without the use of manipulation. But if you can't visualize it, I think chances are you're gonna have to play around with the space for a while. I hope that helps.
