Showing that a ball is a open set (real analysis) The following is an example from my textbook.
Claim: A ball given by $K(a,\rho) = \left\{x \in \mathbb R^k| ||x-a|| < \rho \right\}$ is an open set.
Proof:
The idea is to take an arbitrary $x \in K(a,\rho)$ and show that it is an interior point. If $x$ is an interior point then there exists a small ball $K_s$ around it such that $K_s \subset K(a,\rho)$.
(Here comes the part I do not understand)
Choose an $\epsilon \in (0,\rho-||x-a||)$. If $y \in K(x,\epsilon)$, then from the triangular inequality we get
$$||y-a|| = ||y-x+x-a|| \le ||y-x||+||x-a|| < \epsilon + ||x-a|| \underset{\text{This inequality sign i do not understand}} < \rho$$
Q1: What are the ideas behind the proof?
Q2: What is $\epsilon$? Can it be $0$?
Q3: What is $\rho-||x-a||$ ? A line segment from $x$ to the belonging sphere with radius $\rho$?
PS: I'm translating to English, so bear with me. 
 A: For $Q2:$ No, $\varepsilon$ cannot be $0$. You wrote yourself that $\varepsilon$ lies in the open interval $(0,\rho-\|x-a\|)$. \
For $Q3:$ $\rho-\|x-a\|$ is the distance between $x$ and the boundary of $K(a,\rho)$. Try drawing the situation: $a$, the ball around $a$, your point $x$ and the straight line between $a$ and $x$.
For $Q1:$ Using the above drawing is the idea: The point on the boundary closest to $x$ lies on the straight line between $x$ and $a$.
A: From your question, I can say that you have a deep weakness in visualizing open balls and the topic "Sets in $\Bbb R$". Try to read the basics from Rudin W. Principles of Mathematical Analysis.
For now, I am clearing the doubts.
$1.$ The idea behind the proof: Consider the same problem in $\Bbb R$ with $|\cdot |$ (Mod, which is norm for $\Bbb R$). We need to prove that $(0,1)$ is an open set. We take arbitrary $x\in (0,1)$ then we can find a $\epsilon >0$ for which $x\in (x-\epsilon,x+\epsilon)\subset(0,1)$ which proves $x$ is an interior point of $(0,1)$ and this is true for all $x\in(0,1)$, so $(0,1)$ is an open set. In case of $\Bbb R^n$ we have $||\cdot ||$ and we repeat the process.
$2.$ The $\epsilon$ thing comes from the definition of the interior point of a set.  Let $X$ be a nonempty subset of $\Bbb R^n$, a point $x\in X$ is called an interior point of $X$ if we can find an open ball/open nbd that contains $x$ and contained in $X$ i.e. if we can find a $r>0$ such that $x\in B(x,r)\subset X$, where $B(x,r)=\{y:||x-y||<r\}$.
So basically the $\epsilon$ thing you choose is the radius of the inner sphere, which can not be zero (radius zero means it's singleton set, which is not open).
$3.$ The quantity $\rho-||x-a||$ is a number chosen suitably so that we can fit an open ball of radius $\epsilon$ which is greater than zero but less than $\rho -||x-a||$ inside the set $K(a,\rho)$
The inequality you didn't understand: That's homework! Draw the circles by taking $\rho=2$ cm $a=(0,0)$ (Origin in $\Bbb R^2$). And enjoy the beauty!
A: 
Let's look at the case in $R^2$ first. The idea in the proof is to choose the radius of the ball centered at $x$ such that it is less than $p - \| x-a\|$. As you can see from the picture, that ball will be contained in the big ball, thus making the big ball an open set. When you try to prove that any point in the ball centered at $x$ is in the big ball, you can simply use the triangle inequality as shown in your textbook. The case in $R^n$ is the same; it's just easier to imagine in $R^2$. The inequality you mentioned above is because if you choose a point y inside the ball centered at $x$, then $\| y-x\| < \epsilon  $ due to $\epsilon$ being the radius of the ball.
