Equivalent definition $\text{Cone}(K)$ 
Let $K\subseteq \mathbb{R}^{n} \times \{ 0 \}\subseteq \mathbb{R}^{n+1} $ and $v=(0,0,\cdots,1) \in \mathbb{R}^{n+1}$.
For every $x \in K$,  let $$Lx=\left \{ tv+(1-t)x \; | \ t \in [0,1] \right \}$$
Define 
  $$
\text{Cone}(K)=\bigcup_{x \in K}Lx.
$$
  namely the union of all segments from $v$ to every point of $K$.
According to all definitions that I see on books, Cone is defined: https://en.wikipedia.org/wiki/Cone_(topology)
However the exercise I'm working on  asks to find under what circumstances  $$\text{Cone}(K)=\bigcup_{x \in K}Lx$$ is isomorphic to $(K \times [0,1]) /(K \times \{ 1 \})$ (quotient space).

For example if $$K=\{(1/n,0):n\in\mathbb{N}\} \cup (0,0)$$ I can see both of sets form the same graph but I'm still unable to derive a homeomorphism between these two objets.
 A: Let $f:K\times[0,1]\to \mathrm{Cone}(K)$ be defined by $f(k,t)=(1-t)k+tv$.
Then, $f$ restricted to $K\times\{1\}$ is constant so $f$ yields a continuous map $$(K\times[0,1])/K\times\{1\}\to \mathrm{Cone}(K)$$
It is clearly a bijection, so if $K$ is compact, then it is automatically a homeomorphism, which is the case for your example.
The problem with $K$ not compact is that $\mathrm{Cone}(K)$ has bad properties.
Take $K:=\mathbb R\times\{0\}\subset\mathbb R^2$, then $$\mathrm{Cone}(K)=\mathbb R\times[0,1)\cup\{v\}\subset \mathbb R^2$$
Look at the neighborhoods or $v$ : they do all contain a subset of the form $D(N,\varepsilon)\cap\mathrm{Cone}(K)$ by definition of the subspace topology.
Look at the inverse image of some $D(N,\varepsilon)\cap\mathrm{Cone}(K)$ by $f$ : it is an open neighborhood of $\mathbb R\times\{1\}$ in $(\mathbb R\times[0,1])/(\mathbb R\times\{1\})$. So by definition of the quotient topology, it is the projection of an saturated open neighborhood $U$ of the subset $\mathbb R\times\{1\}$ in $\mathbb R\times[0,1]$. 
For $k=(x,0)\in K$, $|v-f(k,t)|=(1-t)|v-k|=(1-t)\sqrt{1+x^2}$, so $$(x,t)\in U\iff |v-f(k,t)|<\varepsilon \iff t> 1-\frac{\varepsilon}{\sqrt{1+x^2}}$$
so that $U$ is the strict epigraph of the function $\mathbb R\to [0,1]$, $$x\mapsto 1-\frac{\varepsilon}{\sqrt{1+x^2}}$$
We have shown that if $V$ is an open neighborhood of $v$ in $\mathrm{Cone}(K)$, then its reverse image by $f$ contains the image in the quotient $(\mathbb R\times[0,1])/(\mathbb R\times\{1\})$ of the strict epigraph of a function $$x\mapsto 1-\frac{\varepsilon}{\sqrt{1+x^2}}$$ for some $\varepsilon$.
But for $g(x)=1-e^{-|x|}$, the image in $(\mathbb R\times[0,1])/(\mathbb R\times\{1\})$ of the strict epigraph of $g$ is an open set of the topological cone of $K$, which image by $f$ is not open is $\mathrm{Cone}(K)$ by what we have shown, and a classical argument of real analysis. The topology of the topological cone is then finer (has more open sets) than that of $\mathrm{Cone}(K)$.
So the answer is that in general, you can't identify those two spaces. But it works for the compact case.
A: This is a supplement to elidiot's answer.
Let us show that if $K$ is not compact, then the spaces $CK = K \times I/K \times \{ 1 \}$ and $\text{Cone}(K)$ are not homeomorphic (although the map $(K\times[0,1])/K\times\{1\}\to \mathrm{Cone}(K)$ described in elidiot's answer is always a continuous bijection).
The space $\text{Cone}(K)$ is metrizable because it is a subspace of $\mathbb R ^{n+1}$, in particular each point has a countable basis of open neighborhoods. We shall show that the cone point $* \in CK$ (which is the common equivalence class of all points $(k,1) \in K\times\{1\}$) does not have a countable basis of open neighborhoods which implies our above claim.
We argue by contradiction. Assume that $*$ has a a countable basis of open neighborhoods $U_n$.  Let $p : K \times I \to CK$ denote the quotient map. Then the $V_n = p^{-1}(U_n)$ are open neighborhoods of $K\times\{1\}$ in $K \times I$. Since $K$ is not compact, there exists a sequence $(x_n)$ in $K$ which converges to some $\xi \notin K$ (if $K$ is unbounded, we simply take $\xi = \infty$ although this is not a real number). W.l.o.g. we may assume that $x_n \ne x_m$ for $n \ne m$. The set $A = \{ x_n  \mid n \in \mathbb N \}$ and all sets $A_m  = A \setminus \{ x_m \}$ are closed in $K$.
For each $n$ there exists $t_n < 1$ such that $(x_n,t_n) \in V_n$. Now define $C = \{ (x_n,t_n)  \mid n \in \mathbb N \}$. This set is closed in $K \times I$. To see this, consider any sequence $(y_k,s_k)$ in $C$ which converges to some $(y,s) \in K \times I$. This means $y_k \to y$ and $s_k \to s$. Since $(y_k)$ is a sequence in $A$, we have $y = x_m \in A$ for some $m$. This implies that $y_k = x_m$ for $k \ge k_0$ (otherwise we could construct a subsequence $(y_{k_r})$ of $(y_k)$ which is contained in $A_m$ but converges to $x_m \notin A_m$ which is impossible because $A_m$ is closed in $K$). This implies that $(y_k,s_k) = (x_m,t_m)$ for $k \ge k_0$ and therefore $(y,s) \in C$.
We conclude that $V = K \times I \setminus C$ is an open neighborhood of $K\times\{1\}$ in $K \times I$. We have $p^{-1}(p(V)) = V$, thus $p(V)$ is an open neighborhood of $*$ in $CK$. Hence there exists $n$ such that $U_n \subset p(V)$. By construction $p(V) \cap p(C) = \emptyset$. But now
(1) $p(x_n,t_n) \in p(V_n) = U_n \subset p(V)$
(2) $p(x_n,t_n) \in p(C)$
which is a contradiction.
