Proof: Any isomorphism between 2 graphs G and H map each vertex to a vertex of the same degree I wrote a proof by contradiction for the problem in the title:
Reword the statement to "if there exists an isomorphism between graphs G and H, then the isomorphism maps each vertex of G to a vertex of H of the same degree". 
Following a proof by contradiction format, we assume that (1) there exists an isomorphism, and (2) the isomorphism maps vertex x$\in$G to the corresponding vertex x'$\in$H, with d(x) > d(x'). Thus, x$\in$G has at least one more adjacent vertex than x'$\in$H, we'll pick one of those adjacent vertices and label them "a".
There exists an edge x - a$\in$G such that there is no corresponding edge in Y with x' as an end, since d(x) > d(x'). This contradicts the hypothesis, which states that there is indeed such a corresponding edge, and we are done the proof.
I don't feel too confident about my work; I don't think it's rigorous/"solid" enough. Any advice on how to improve it would be appreciated.
 A: You are picking $a$ as a vertex adjacent to $x$, but later it must have the
property that the edge $xa$ does not correspond to an edge in $H$.
How do you know that such an $a$ exists?
Let's write $a_1,\ldots,a_m$ as the vertices adjacent to $x$ in $G$,
where $m=d(x)$, and let $a_1',\ldots,a_m'$ be the vertices in $H$
corresponding by the given isomorphism to $a_1,\ldots,a_m$. Since the
isomorphism is an isomorphism, each of $a_1',\ldots,a_m'$ neighbour
$x'$. What does that tell you about $d(x')$?
A: This theorem follows simply from the definition of isomorphic graphs presented here. Namely, two graphs $G$ and $H$ are called isomorphic if there exists a bijection $p : V(G)\to V(H)$ such that for $u, v\in V(G)$, $\{u, v\}\in E(G)$ with multiplicity $n$ iff $\{p(u), p(v)\}\in E(H)$ with multiplicity $n$. Then, $p$ is called an isomorphism between $G$ and $H$. Here, $V(G)$ and $E(G)$ are the set of vertices and the multiset of edges of graph $G$, respectively.
Thus, we let $p$ be an isomorphism between $G$ and $H$. Given a vertex $u\in V(G)$, we let $E_u(G)\subset E(G)$ be the multiset of edges incident to $u$ (with loops counted twice), so $\deg(u) = \lvert E_u(G)\rvert$. Then, it's pretty easy to see that the map $\{u, v\}\mapsto \{p(u), p(v)\}$ is a bijection between $E_u(G)$ and $E_{p(u)}(H)$ (this follows directly from our definition of isomorphism on graphs), so we will have $\deg(u) = \deg(p(u))$.
