Prove that a tree with a vertex $v$ of degree $k > 1$ has at least $k$ leaves

Your introduction of $v'$ is a little confusing. I would instead say "Let $v'$ be a vertex of maximal degree in $T$. Since we know there is a vertex of degree $k>1$ in $T$, it follows that $\delta_T(v') \geq k$." Or alternatively, use some notation in addition to words, such as "There is a vertex $v'$ such that $\delta_T(v') \geq \delta_T(v'')$ for any vertex $v''$ of $T$. Since $\delta_T(v)=k$, it follows that $\delta_T(v'') \geq k$."

Note that you can't assume $\delta_T(v')=k$ only that $\delta_T(v') \geq k$, but that does not affect the inequality later in your proof.

As an aside, I don't think the restriction $k>1$ is necessary. If $k=1$ then the only possibility is the path graph with two vertices, which has at least $1$ leaf.

It is a nice complete proof, but what about the sentence " This means that there is a vertex that has more than one edge and also has an equal or greater number of edges than any other T vertex."? The vertex $v'$ need not be the vertex with the maximum degree, and the proof will still be right.

Also the notation $\delta(v)$ is usually used to denote the minimum degree in a graph, so I think it is better to use $d(v)$