Proof that a graphic sequence realizes a specific graph structure I have a problem understanding not only this proof, but also the theorem.
Definition:   A non-increasing sequence $\langle d_{1}, d_{2}, ..., d_{n} \rangle$ is said to be graphic if it is the degree sequence of some simple graph.
Theorem: Let $\langle d_{1}, d_{2}, ..., d_{n} \rangle$ be a graphic sequence, with $d_{1} \geqslant d_{2} \geqslant ... \geqslant d_{n}$. Then there is a simple graph with $\{v_{1}, v_{2}, ..., v_{n}\}$ as a vertex-set satisfying $deg(v_{i}) = d_{i}$ for $i = 1, 2, ..., n$, such that $v_{1}$ is adjacent to vertices $v_{2}, ..., v_{d_{1} + 1}$.
I don't understand the last part of the theorem. Does this mean that the vertex with the highest degree (for example 5) is adjacent to the next 5 vertexes with the highest degrees in the sequence (for example 4, 3, 2, 2, 1, but not 2, 2, 2, 2, 1)?
Proof:
Among all simple graphs with $\{v_{1}, v_{2}, ..., v_{n}\}$ as a vertex-set and $deg(v_{i}) = d_{i}, i = 1, 2, ..., n$, let $G$ be one for which $r = | N_{G}(v_{1}) \cap \{v_{2}, ..., v_{d_{1} + 1}\}|$ is maximum [1].
If $r = d_{1}$, then the conlusion follows. If $r < d_{1}$, then there exists a vertex $v_{s}$, where $2 \leqslant s \leqslant d_{1} + 1$, such that $v_{1}$ is not adjacent to $v_{s}$, and there exists a vertex $v_{t}$, where $t > d_{1} + 1$ such that $v_{1}$ is adjacent to $v_{t}$, since $deg(v_{1}) = d_{1}$ [2].
Moreover, since $deg(v_{s}) \geqslant deg(v_{t})$, there exists a vertex $v_{k}$ such that $v_{k}$ is adjacent to $v_{s}$ but not to $v_{t}$. Let $\tilde{G}$ be the graph obtained from $G$ by replacing the edges $\{v_{1}, v_{t}\}$ and $\{v_{s}, v_{k}\}$ with the edges $\{v_{1}, v_{s}\}$ and $\{v_{t}, v_{k}\}$. Then the degrees are all preserved and $v_{s} \in N_{\tilde{G}}(v_{1}) \cap \{v_{3}, ..., v_{d_{1} + 1}\}$.
Thus, $|N_{\tilde{G}}(v_{1}) \cap \{v_{2}, ..., v_{d_{1} + 1}\}| = r + 1$, which contradicts the choice of $G$ and completes the proof [3].
Questions:
[1]: What does that mean exactly, in simple words? Does that mean that it is a graph, where the vertex with the highest degree is connected to other vertices with highest degrees? This is the same problem that I have with the theorem I guess.
[2]: Why does $v_{t}$ have to have $t > d_{1} + 1$ and why does it have to be adjacent to $v_{1}$?
[3]: I am not sure I understand the last step. However, I think that it would make sense once I understand the previous ones.
Thank you for the clarifications.
 A: Yes, the goal is to get a graph where the highest-degree vertex $v_1$ is adjacent to the $d_1$ next-highest degree vertices.
This has no purpose of its own, but it's important because this is the one thing we know we can always guarantee. The degree sequence $d_1, \dots, d_n$ might have many realizations, and $v_1$ might have different sets of neighbors in them, and we don't know exactly which sets of neighbors are possible. However, we know that the set $\{v_2, \dots, v_{d_1+1}\}$ must be possible.
Later on, this gives us an effective algorithm for checking if a degree sequence is graphic.

The intuitive idea in the proof is that we can start with any graph with this degree sequence, and perform edge swaps (replacing edges $uv$ and $xy$ with edges $ux$ and $vy$) until it has the form required by the theorem.
How can we know we are making progress in the right direction? Well, we can measure "how good" a graph is for our purposes by seeing how many edges $v_1$ has to the vertices in $S = \{v_2, v_3, \dots, v_{d_1+1}\}$. Eventually, we'd like the answer to be "$d$: all of them". We can make sure that we're making progress towards that by proving the following lemma.
Lemma. If $v_1$ is not adjacent to all of the vertices in $S$, then we can perform an edge swap to increase the number of neighbors $v_1$ has in $S$.
Proof. This is what's going on in the second and third paragraphs of your proof. Let me explain more clearly where $v_s, v_t, v_d$ come from.

*

*If $v_1$ is not adjacent to all the vertices in $S$, then there must be some vertex in $S$ that it isn't adjacent to: that's $v_s$.

*In total, $v_1$ must have $d_1$ neighbors. There are $d_1$ vertices in $S$, but $v_1$ isn't adjacent to all of them. So $v_1$ needs a neighbor outside $S$ to get to $d_1$ neighbors: that's $v_t$.

*Since $\{v_1\} \cup S$ has all the highest-degree vertices, we have $\deg(v_s) \ge \deg(v_t)$. Moreover, $v_t$ is at a disadvantage: it is adjacent to $v_1$, but $v_s$ isn't. To make up for that, there must be at least one vertex adjacent to $v_s$ but not $v_t$: that's $v_k$.

At this point, we do what your proof does: delete $v_1v_t$ and $v_s v_k$, then add $v_1 v_s$ and $v_t v_k$. (All this work was to make sure that edges $v_1 v_s$ and $v_t v_k$ don't yet exist.) Now $v_1$ has one more neighbor in $S$: the vertex $v_s$.

Now I can explain why we say "Let $G$ be the graph for which $|N(v_1) \cap S|$ is maximum".
This is something called the extremal principle. It's a shortcut for proof-writing. We have a lemma that says: if $N(v_1) \ne S$, it's possible to change the graph to make $|N(v_1) \cap S|$ larger. Intuitively, what we want to say is "apply this lemma until you can't apply it anymore", but that sounds vague.
So instead, we pick the graph $G$ with the largest possible value of $|N(v_1) \cap S|$, because then we know that the value of $|N(v_1) \cap S|$ can't be increased any further. The lemma says "it can always be increased further, with one exception: if $N(v_1) = S$". So we conclude that $G$ has $N(v_1) = S$.
