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?

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 $V\cap \varphi(X)=U\times \{t\}=\varphi(U).$
