For fixed $t$ with $0 \le t < 1$, prove that $x \mapsto [x, t]$ defines a homeomorphism from a space $X$ to a subspace of the cone $CX$. 
For fixed $t$ with $0  \le t < 1$, prove that $x \mapsto [x, t]$ defines a homeomorphism from a space $X$ to a subspace of the cone $CX$.

Let $\varphi : X \to \varphi(X)$ be the map defined by $x \mapsto [x,t]$. Then $\varphi$ is clearly surjective as the codomain is defined to be the image of it. To prove that $\varphi$ is injective I considered the standard approach to assume that $\varphi(x)=\varphi(y) \implies [x,t]=[y,t]$, but I don't think I can conclude from here that $x=y$? I am also wondering what is the idea with $t <1$ being a strict inequality? Are we excluding the vertex of the cone here, if so why?
 A: As discussed in the comments, the fact that $\varphi : X \to \varphi(X)$ is injective follows from the definition of the equivalence relation on the cone, since for $t < 1$ the equivalence classes are singletons, that is $[x,t] = \{(x,t)\}$ for every $t \in [0,1)$ and $x \in X$.
Now, for the continuity we need to look at open sets in $\varphi(X)$ and check that their preimages are open in $X$. By the definition of the quotient topology induced by an equivalence relation $\sim$ on a space $W$, a set $U \subset W/\sim$ is open if and only if the set $\{w \in W \; | \; [w] \in U\}$ is open in $W$.
In our case $W = X \times [0,1]$ and $\sim$ is the relation on $X \times [0,1]$ defined by $(x,1) \sim (y,1)$ for every $x, y \in X$ (hence $W/\sim = CX$). Thus, a set $U \subset CX$ is open if and only $\{(x,t) \in X \times [0,1] \; | \; [x,t] \in U\}$ is open in $X \times I$.
Since we are only interested in subsets of $CX$ which do not contain the vertex, we know that $[x,t] = \{(x,t)\}$ for every equivalence class $[x,t]$ in our subset. This implies that a subset $U$ (of the cone without the vertex) is open if and only if $\{(x,t) \in X \times [0,1) \; | \; \{(x,t)\} \in U\}$ is open in $X \times [0,1)$, that is if $U$, seen as a subspace of $X \times [0,1)$ is open in $X \times [0,1)$ (since the subset of the cone without the vertex is "the same" (up to renaming $[x,t]$ to $(x,t)$) as the subset of the cylinder without its upper base).
Fixing $t = t_0$ and using the definition of open sets of the subspace topology, we see that a subset $U$ of $\varphi(X)$ is open if and only if $\{(x,t_0) \in X \times \{t_0\} \; | \; [x,t_0] \in U\}$ is open in $X \times \{t_0\}$.
Now you can conclude with the definition of the open sets in the product topology $X \times \{t_0\}$.
A: Drawing a picture helps. Basically you're just trying to show that slices of the cylinder embed into the cone as long as $t\neq 1.$
Fix $0\le t<1.$ Then, $[x,t]\sim [y,t]\Leftrightarrow  x=y$ because by definition of $\sim$, the equivalence classes are singletons when $0\le t<1.$ You have shown that $\varphi $ is continuous. So all you need to show is that it is an open map $\textit{onto}$ its image, $\varphi(X)$ with the $\textit{subspace}$ topology of the quotient topology on $C(X).$
Therefore, let $U\subseteq X$ be open. We want to show that $\varphi (U)$is open in $\varphi (X).$ This means we should find an open set $V$ in $C(X)$ such that $V\cap \varphi(X)=\varphi(U).$
But this is easy. Note that $V=U\times [0,1)$ is an open set in $C(X).$ And $V\cap \varphi(X)=U\times \{t\}=\varphi(U).$
