I've come up with an answer to this problem that I have precisely zero faith in, so a 'yay' or 'nay' would be enormously appreciated with regards to whether I'm an idiot or not. So here it goes:
$(X,\mathcal{T})$ is a topological space, and $\mathbb{I}= \{t \in \mathbb{R} \;\; | \;\; 0 \leq t \leq 1 \}$ has the Euclidean topology. $X \times \mathbb{I}$ is given the product topology, and for $\lambda \in \mathbb{I}$, take the function $i_\lambda: X \longrightarrow X \times \mathbb{I}, \quad x \longmapsto (x,\lambda)$ The image of $i_\lambda$, $\text{im}(i_\lambda)$, has the topology induced by the product topology on $X\times\mathbb{I}$. I need to show that $\phi_\lambda: X \longrightarrow \text{im}(i_\lambda),\quad x\longmapsto i_\lambda(x)$ is a homeomorphism.
OK, so, firstly I decided that $\text{im}(i_\lambda) = X \times \{\lambda\}$.
For injectivity, given $a,b \in X$, $\phi_\lambda(a) = \phi_\lambda(b) \Longleftrightarrow (a,\lambda)=(b,\lambda) \Longleftrightarrow a=b$.
For surjectivity, given $(a,\lambda)\in X\times \{\lambda\}$, $\exists b\in X$ such that $\phi_\lambda(b) = (a,\lambda)$, namely $b=a$.
So $\phi_\lambda$ is bijective.
Now I define $\zeta_\lambda: X\times \{\lambda\} \longrightarrow X,\quad (x,\lambda)\longmapsto x.$ I think that $\zeta_\lambda \circ \phi_\lambda = \text{id}_X$ and that $\phi_\lambda \circ \zeta_\lambda = \text{id}_{X\times \{\lambda\}}$, so $\zeta_\lambda = \phi_\lambda^{-1}$. Now it just remains to be shown that $\phi_\lambda$ and $\phi_\lambda^{-1}$ are continuous. To this end, I showed that the topology on $\text{im}(i_\lambda)=X\times\{\lambda\}$ is
$\mathcal{G} = \{U\times\{\lambda\}\;|\; U \in \mathcal{T}\,\},$ and so to show that $\phi_\lambda$ is continuous, take $A \in \mathcal{G}$, i.e. $A=U\times\{\lambda\}$ for some $U\in \mathcal{T}$. Then $\phi_\lambda^{-1}[A] = \phi_\lambda^{-1}[U\times\{\lambda\}]=U\in\mathcal{T}.$ Thus $\phi_\lambda$ is continuous. Then for $\phi_\lambda^{-1}$, take $K\in\mathcal{T}$; $\left(\phi_\lambda^{-1}\right)^{-1}[K] = \phi_\lambda[K] = K\times\{\lambda\} \in \mathcal{G}.$ Hence $\phi_\lambda^{-1}$ is continuous.
Therefore $\phi_\lambda$ is a homeomorphism.
Please be as brutally honest as you feel you need to be, as I have no delusions about my mathematical ability. If it makes no sense and I understand nothing, so be it. Thank you for your time.