Defining $T$: Given a connected graph $G$ with $n$ vertices, let $T(G)$ count the number of sequences $$x_{1},x_{2},...,x_{n}$$ of the vertices of $G$ such that the induced subgraphs $$G_{i}=G[x_{1},x_{2},...,x_{i}]$$ are connected.
Example: Take the cyclic graph $C_{4}$. $T(C_{4})=4*2^{4-2}=16$ since we start with 4 vertices to choose from (meaning if removed, the graph is still connected), then 2 (either of the 1-connected vertices), then 2 again (at this point the graph has 2 vertices and 1 edge; we count the graph of a single point).
What is known: $G$ is a connected graph. $H$ is a subgraph of $G$, and $H$ is connected.
What to show: $T(H)\le T(G)$.
Work done: To be honest, I don't even know where to start. If the subgraph was induced then I could make an argument considering the induced edges, but that is not specified in the problem statement so if I use that, I only considered half of the proof, at least I believe. I would appreciate any help.
Given a subgraph $H$ of $G$, let $K$ be the induced version of $H$. This means $K$ has the same vertex set as $H$, and all induced edges from $G$. You have proved that $T(K)\le T(G)$.
I claim that $T(H)\le T(K)$. Indeed, since $K$ as all the edges that $H$ has, for any sequence of vertices $x_1,\dots,x_k$ of $K$. we have $$ \forall i\quad H[x_1,\dots,x_i]\text{ is connected} \quad\implies\quad K[x_1,\dots,x_i]\text{ is connected} $$ This is true since adding edges to a connected graph preserves connectedness.
It follows that every valid sequence for $T(H)$ is one for $T(K)$, and we conclude $T(H)\le T(K)\le T(G)$.