Show $T$ is a homeomorphism from $E$ onto $T(E)$ Let $E$ be the open interval $(0, 1)$ with the usual metric
topology and $\Bbb R$ be given the usual topology.
Let $\Bbb R^J$ be given the topology of pointwise convergence $T_p$, where the index set $J$ comprises all continuous functions $f : E \to \Bbb R$. 
Define a function $T : E \to \Bbb R^J$ by $T(x) = (f(x))_{f∈J}$ . Suppose that the image set $T(E)$ is given the subspace topology as a subspace of $(\Bbb R^J, T_p)$.
I want to show that $T$ is a homeomorphism from $E$ onto $T(E)$ but am not sure what method to employ even hmm. 
Is there an explicit map to find? or find continuous maps $f$ and $g$ s.t. $f\circ g=g\circ f= i$ where $i$ is the identity map? 

You have the map $T$. This is the homeomorphism between $E$ and and $T[E]$.
So you need to show that $T$ is 1-1/injective. This comes down to the fact that if $x \neq y$ in $E$ there is a continuous function $f: E \to \mathbb{R}$ (so $f \in J$) such that $f(x) \neq f(y)$.
Then $T$ is shown to be a bijection between $E$ and $T[E]$.
$T$ is continuous because $\pi_f \circ T = f$ is continuous for every $f \in J$. 
$T$ has a continuous inverse too. This is the trickest part. This turns out to be equivalent to the fact that if $C \subseteq E$ is closed and $x \notin C$, there is a function $f \in J$ such that $f(x) \notin \overline{f[C]}$. But in this case you might be able to find a more specific direct argument, if you look for it
But this argument in fact works not just for $E$ but for any completely regular $T_1$ space: these all embed into a power of reals.
