In the first graph, no more edges can be added, because the pairs $(v_1,v_2)$ and $(v_1,v_4)$ have degree sum $3$, and every other pair is already adjacent. So this graph is its own closure.
For the second graph, we can add edges $v_1v_3$ and $v_2v_4$ immediately, since for both pairs the degree sum is $4$. Now the degree sum of $v_1,v_4$ is $5$, so we can add this edge too (note that we could not add this edge initially). There are no more non-adjacent pairs, so we've reached the closure.
The point of this definition is that a simple graph is Hamiltonian if and only if its closure is Hamiltonian - this is the Bondy-Chvátal theorem.
Let $G$ be a $d$-regular bipartite graph. And let $e_{uv}$ be an edge of $G$.
Let $S$ the set of $e_{uv}$ and all its adjacent edges. Then the size of $S$ is $2d-1$ : They are $d$ edges connected at the vertex $u$, and $d$ connected at vertex $v$. As we double count $e_{uv}$, we reduce the count by one, and $|S|=2d-1$.
Recall that an antimatching $A$ of $G$ is a set of edges such that two edges are at most at distance 2, and that the distance between two edges is defined as the distance between corresponding vertices in $L(G)$ (the Line graph of $G$).
As all edges from $S$ are connected to $e_{uv}$, then the maximum distance between edges of $S$ is 2.
Therefore $S$ is an antimatching of size $2d-1$.
I think that's all to proove. (I never studied antimatching before, I might be missing something).
Note An antimatching can also be defined (equivalently) as a subgraph with no induced matching of size greater than 1. This is also clearly the case for $S$ :
Suppose there exist an induced matching of size 2 in $S$. Then it must contains one edge connected to $u$ and one connected to $v$. But then in order to be induced we would need to include $e_{uv}$, and would not be a matching. Hence a contradiction.
Therefore the only induced matching of $S$ have size 1.
Best Answer
An example should help.
Think about the triangle as a graph with three vertices $A, B, C$ and three edges $(AB), (BC), (AC)$. Then the total graph of $G$ has six vertices: $$ A, B, C, (AB), (BC), (AC) . $$ Its edges are the pairs of vertices $$ (AB), (BC), (AC) $$ (corresponding to adjacent vertices of $G$), $$ ((AB)(AC)), ((AB)(BC)), ((AC)(BC)) $$ (corresponding to edges that share a vertex) and edges like $$ (A, (AB)) $$ (corresponding to vertices on edges).