A valid tree decomposition of a graph $G =(V,E)$ is also a valid tree decomposition of any subgraph $G' = (V', E') $ provided that you simply remove all vertices $v \in V \setminus V'$ from all the bags.
Let's look at the definition. A tree decomposition $\mathcal T$ of $G$ is a tree of "bags" where each bag is a subset of vertices from $G$ restricted to the following three constraints:
- The union of all the bags equals $V(G)$
- Every edge $e \in E(G)$ is in some bag of $\mathcal T$, i.e. if $e = uv$, then there is a bag $B_i \in V(\mathcal T)$ such that $\{u,v\} \subseteq B_i$, and
- for every vertex $v \in V(G)$, $\mathcal T_v$ the subgraph consisting of all bags of $\mathcal T$ containing $v$ is connected, i.e., $\mathcal T_v$ is itself a tree.
Let $\mathcal T$ be a tree decomposition of $G$, let $G' = (V', E')$ be a subgraph of $G$, and let $\mathcal{T'}=\mathcal{T}_{|V'}$ be the tree $\mathcal T$ but where you delete all vertices not in $V'$.
Clearly (1) the union of the bags in $\mathcal{T}'$ is $V'$, (2) every edge $uv \in E'$ is in some bag in $\mathcal{T}'$ since $\{u,v\} \subseteq V'$ and therefore they must be in some bag in $\mathcal{T}$, and (3), the connectivity constraint holds for $v \in V'$ since it holds in $\mathcal{T}$.