Minimum spanning tree and its connected subgraph
Flaws in the accepted answer: When I re-read the accepted answer given by @Mahmoud A. today, I find flaws in it.       

Consider the figures for an example: 


In this example, $e = CE$. In figure (5), the edge $CD$ in the cycle created by adding edge $e = CE$ into $T_H$ is not in $T_G$ shown in figure (2), but we have $w(CD)&lt;w(e=CE)$. So the claim "For every other edge $e' \notin T_G$ in the cycle we have $w(e′)=w(e)$" does not hold.

Note: In the following, I fix the flaws. I leave the answer given by @Mahmoud A. as accepted because it is original and insightful.

Correction: While where is an edge $e \in T_G \cap H$ such that $e \notin T_H$ do:


Adding $e$ to $T_H$ to create a cycle $C$.      


1.1 Because $T_H$ is an MST of $H$, for every other edge $e' \in C$ we have $w(e') \le w(e)$. Otherwise, for any edge $e' \in C$ such that $w(e')&gt;w(e)$, $T_H + \{ e \} - \{ e' \}$ has smaller weight than $T_H$.
1.2 Let $e = (u,v)$. Remove $e$ from $T_G$. Then vertices $u$ and $v$ are separated into two different connected components, denoted by $U \ni u$ and $V \ni v$, of $T_G$. Because $T_H$ is an MST of $H$ (which includes vertices $u$ and $v$) and $e \notin T_H$, there must be an edge $e'' \in C$ which connects the two components $U$ and $V$ again.
We claim that $w(e'') = w(e)$. In 1.1, we have proved that $w(e'') \le w(e)$ (replacing $e'$ there by $e''$). If $w(e'') &lt; w(e)$, then $T_{G} - \{ e \} + \{ e'' \}$ has smaller weight than $T_G$.      
1.3 According to 1.1 and 1.2 above, there are an edge $e'' \in C$ such that $e'' \neq e \land w(e'') = w(e)$. We replace $e''$ with $e$ in $T_H$ to obtain $T'_H = T_H - \{ e'' \} + \{ e \} \in MST_H$.
Since $e'' \notin T_{G} \cap H$ and $e \in T_{G} \cap H$, $T'_H$ contains one more common edge with $T_{G} \cap H$ than $T_H$ does.

Rename $T'_H$ to $T_H$


EndWhile

After the loop we have $T_{G} \cap H \subseteq T_H$.