Let G be a 6-regular graph of order 10 and let $u, v \in V(G)$. I want to prove that $G$, $G-v$ and $G-v-u$ are all Hamiltonian.
This is quite interesting. Proving G is Hamiltonian is rather easy. The sum of degrees of any two non-adjacent vertices of G is 12, which is greater than 10, making G Hamiltonian. $G-v$ has 6 vertices of degree 5 and 3 of degree 6, which still always add up to at least 10, which guarantees $G-v$ is indeed Hamiltonian.
There are some problems with this approach, though. The first and most obvious is that applying it to $G-v-u$ doesn't provide credible guarantees it is Hamiltonian. The second is that it's less elegant.
I thought of using the Bondy-Chvatal Closure Theorem because is seems fitting, but I can't exactly figure out how to do so. Any help is appreciated, as I'm an economist by training and math isn't particularly my forte.
You are right to use Ore's theorem: if $G$ is a graph on $n\geq 3$ vertices and $\deg(x)+\deg(y)\geq n$ for all $xy\notin E$, then $G$ is hamiltonian. You've correctly applied this to $G$ since $\deg(x)+\deg(y)=12\geq 10$ for all $x,y\in V$. You've also applied this to $G-u$ since $\deg(x)+\deg(y)\geq 5+5=10\geq 9$ for all $x,y\in V(G-u)$ ($G-u$ has $9$ vertices!) For $G-u-v$, each vertex has degree at least $4$ and $G-u-v$ has $8$ vertices, so certainly $\deg(x)+\deg(y)\geq 8$.
Really we're not using the full power of Ore's theorem, but instead just using Dirac's theorem, which says that if $G$ is a graph on $n\geq 3$ vertices and $\delta(G)\geq n/2$, then $G$ is hamiltonian. Note that this is the case for $G$, $G-u$ and $G-u-v$.