How to exhibit the set of all the limit points of this subset of $\mathbb{R}^k$? Let $k$ be a positive integer, let $p_0$ be a point in $\mathbb{R}^k$, let $\delta_0$ be a positive real number, and let the set $E$ be defined as follows: 
$$E \colon= \{ \, p\in\mathbb{R}^k \, \colon \, \Vert p - p_0 \Vert < \delta_0 \, \}. $$ Here $\Vert \cdot \Vert$ denotes the Euclidean norm. 
Then how to rigorously show, using the definition of a limit point, that 
$$E^\prime = \{ \, p\in\mathbb{R}^k \, \colon \, \Vert p - p_0 \Vert \leq \delta_0 \, \}? $$ Here by $E^\prime$, I mean the set of all the limit points in $\mathbb{R}^k$ of the set $E$. 
First of all, we have to show that, for every point $p\in\mathbb{R}^k$ such that $\Vert p-p_0 \Vert \leq \delta_0$ and for every real number $\delta > 0$, the intersection 
$$E \cap \left(\mbox{N}_\delta(p)-\{p\}\right) \neq \emptyset,$$ 
where $\mbox{N}_\delta(p)$ is defined as the $\delta$-neighborhood of the point $p$; that is, 
$$\mbox{N}_\delta(p) \colon=  \{ \, x\in\mathbb{R}^k \, \colon \, \Vert x-p \Vert < \delta \, \}.$$ It is this that I'm finding myself unable to show. I mean I'm not managing to explicitly determine an element in the above intersection in terms of the given points $p_0$ and $p$ of $\mathbb{R}^k$ and also in terms of the real numbers $\delta$ and $\delta_0$. 
Then we have to show that for any point $q \in \mathbb{R}^k$ such that $\Vert q-p_0\Vert > \delta_0$, we can find a real number $\delta_q > 0$ such that 
$$\mbox{N}_{\delta_q}(q) \cap E = \emptyset.$$ 
If $x \colon= (x_1, \ldots, x_k) \in \mathbb{R}^k$, then we define $\Vert x \Vert$ as follows: 
$$\Vert x \Vert \colon= \sqrt{x_1^2 + \cdots + x_k^2}. $$
 A: Suppose that $\|p-p_0\|\le\delta_0$, and $\delta>0$; we want to find a point $q$ such $\|q-p_0\|<\delta_0$ and $0<\|q-p\|<\delta$. Assume for now that $p\ne p_0$; we’ll take care of the special case $p=p_0$ at the end. A good place to look for $q$ is on the line segment $\overline{p_0p}$, which consists precisely of the points $tp_0+(1-t)p$ such that $0\le t\le 1$: as $t$ runs from $0$ to $1$, $tp_0+(1-t)p$ runs along the segment from $p$ to $p_0$. 
For $0\le t\le 1$ let $q(t)=tp_0+(1-t)p$; then $q\left(\frac14\right)$ is a quarter of the way from $p$ to $p_0$, $q\left(\frac13\right)$ is a third of the way from $p$ to $p_0$, and so on. Thus, we expect that $\|q(t)-p_0\|<\delta_0$ for $0<t\le 1$, and we expect to be able to make $\|q(t)-p\|$ small but non-zero by taking $t$ small but positive.
In fact, $\|q(t)-p_0\|=\|(t-1)p_0+(1-t)p\|=(1-t)\|p-p_0\|\le(1-t)\delta_0$, so we do indeed have $\|q(t)-p_0\|<\delta_0$ for $0<t\le 1$. Moreover, $\|q(t)-p\|=\|tp_0-tp\|=t\|p-p_0\|$, so we can ensure that $\|q(t)-p\|<\delta$ by taking $t<\frac{\delta}{\|p-p_0\|}$. In particular, the point $$q=q\left(\frac{\delta}{2\|p-p_0\|}\right)$$ has the desired properties.
Now suppose that $\|q-p_0\|>\delta_0$. Let $\delta_q=\|q-p_0\|-\delta_0$, and suppose that $x\in N_{\delta_q}(q)\cap E$. Then by the triangle inequality $\|q-p_0\|\le\|q-x\|+\|x-p_0\|<\delta_q+\delta_0=\|q-p_0\|$, which is impossible. Thus, $N_{\delta_q}(q)\cap E=\varnothing$. 

In fact the closed balls of radius $\delta_0$ around $p_0$ and $\delta_q$ around $q$ are tangent at the unique point of $\Bbb R^k$ whose distance from $p_0$ is $\delta_0$ and whose distance from $q$ is $\delta_q$. This is the point $$\frac{\delta_q}{\delta_0+\delta_q}p_0+\frac{\delta_0}{\delta_0+\delta_q}q\;,$$ i.e., the point $tp_0+(1-t)q$ with $t=\frac{\delta_q}{\delta_0+\delta_q}$. This may help you to visualize the geometry.

If $p=p_0$, let $\epsilon=\frac12\min\{\delta_0,\delta\}$, and let 
$$q=p_0+\langle\epsilon,\underbrace{0,0,\ldots,0}_{k-1}\rangle\;;$$
clearly $0<\epsilon=\|q-p\|<\delta$ and $\|q-p_0\|=\epsilon<\delta_0$.
A: In a metric space, the set of limit points of a set $S$ is the set of points $ClS$ that are at distance $0$ from $S$.
Consider $x$ with $d(x,S) \geq \epsilon $. Then $B(x, \epsilon/2)$ is a ball containing $x$ that does not intersect $S$, so $x$ cannot be in the closure of $S$.
Now, if there exists no such $\epsilon$ for $x$, this means every ball $B(x, \epsilon)$ intersects $S$, for all $\epsilon >0$.
A: Let $x_n$ be a sequence of points in $E$, converging to some point $x$. We want to show that $\|x - p\|\leq \delta_0$. 
Assume otherwise; then $x \in \{p : \|p - p_0\| > \delta_0\}$ which is open. Then, there is $\epsilon > 0$ so that $N_\epsilon(x) \cap E' = \emptyset$. In other words, for every $p \in E, \|x-p\| \geq \epsilon$. In particular, $\|x-x_n\| \geq \epsilon$ for all $n$, and so we have a contradiction.
A: We may assume $p_0=0$, whence
$$E=\{p\in{\mathbb R}^k\>|\>\|p\|<\delta_0\}\ .$$
Put
$$E^?:=\{p\in{\mathbb R}^k\>|\>\|p\|\leq\delta_0\}\ .$$
(a) Consider an arbitrary point $p\in E^?$, and let a $\delta>0$ be given. When $p=0$ put $\rho:={1\over2}\min\{\delta,\delta_0\}>0$. Then the point $p':=(\rho,0,\ldots,0)\in{\mathbb R}^k$ lies in $\bigl(N_\delta(p)\setminus\{p\}\bigr)\cap E$. When $p\ne0$ put $\epsilon:={\delta\over 2\|p\|}>0$. Then the point $p':=(1-\epsilon) p$ lies in $\bigl(N_\delta(p)\setminus\{p\}\bigr)\cap E$. – It follows that $E^?\subset E'$.
(b) For the converse consider a point $p\notin E^?$. Then $\delta:=\|p'\|-\delta_0>0$, and the triangle inequality then implies that the neighborhood $N_{\delta/2}(p)$ does not intersect $E$. This implies $p\notin E'$, so that we obtain $E'\subset E^?$.
Together we obtain $E'=E^?$.
