How to prove that the graph constructed below is still $k$-connected I have often seen the following  construction for $k$-connected graphs in some papers (especially involving planar graphs).
Let $G$ be a graph. Let $x$ be a vertex of $G$.  The Neighbourhood of the vertex $x$, denoted by $N_G(x)$, is the set of all vertices which are adjacent to $x$.
Construction.
Step 0: Let $G$ and $H$ be two disjoint $k$-conneced graphs.  Let $u$ and $v$ be two vertices of $G$ and $H$ with same vertex-degree $l$, respectively. Clearly, $l\ge k$. Let $N_G(u)=\{u_1,u_2,\cdots, u_l\}$ and $N_H(v)=\{v_1,v_2,\cdots, v_l\}$.
Step 1: Obtain $G'$ from $G$ by deleting the vertex $u$ and obtain $H'$ from $H$ by deleting the vertex $v$.
Step 2: Add edges $u_iv_i$ for all $1\le i\le l$ in $G'\cup H'$.
We denote by  $F$ the final resulting graph from above steps.
For example, let both $G$ and $H$ be icosahedral graphs. The following figure shows the construction process of $F$.

I am more convinced that $F$ is still $k$-connected. But I have not rigorously proved this assertion. I belived that  Menger's theorem, Expansion Lemma, or Fan Lemma will work.
Menger's theorem. A graph $G$ is $k$-connected if and only if every two vertices are connectedby at least $k$ independent paths.
Expansion Lemma. If $G$ is a $k$-connected graph, and $G_0$ is obtained from $G$ by adding a new vertex $y$ with at least $k$ neighbors in $G$, then $G_0$ is $k$-connected.
$U$-fan: Given a vertex $x$ and a set $U$ of vertices, an $x,U$-fan is a set of $x,U$-paths such that any two of these paths have only $x$ in common.
Fan Lemma (Dirac, 1960) A graph $G$ is $k$-connected if and only if it has at least $k + 1$ vertices and, for every choice of $x,U ⊂ V(G)$ with $|U| ≥ k$, it has an $x,U$-fan of size $k$.
For two vertices $x$ and $y$ in $F$ such that $\{x,y\}\cap\{x,v\}=\emptyset$, if all independent paths in $G\cup H$ between  $x$ and $y$ lies in $G- \cup \{ vv_i | 1\le i\le k\}$ or $H- \cup  \{vv_i| 1\le i\le k\} $ , then clearly $u$ and $v$ also have $k$ independent paths in $F$.  But what about the $k$ independent paths in $G\cup H$ between $x$ and $y$ using some edges $vv_i$ or $uu_i$, or the vertex $u$ or $v$?  I am looking for a rigorous proof.
 A: Briefly my reasoning for proving that the graph $F$ is $k$-connected.
I use all of your notations.
Let $X\subset V(F)$ and $|X|=k-1$.
The following two cases are possible:

*

*$X\cap V(G')\neq\varnothing$ and $X\cap V(H')\neq\varnothing$;

*$X\cap V(H')=\varnothing$.

The case when $X\cap V(G')=\varnothing$ is similar to case 2.
Сase 1. Let us denote by
$Y=X\cap V(G')$ and $Z=X\cap V(H')$.
Since $|Y|\leq k-2$ and $|Z|\leq k-2$, the graphs $G'-Y$ and $H'-Z$ are connected.
Moreover, there exists $i$ such that $u_i,v_i\notin X$. Therefore edge $u_iv_i$ connects graphs $G'-Y$ and $H'-Z$. So the graph $F-X$ is connected.
Case 2. Let $W$ be one of the connected components of graph $G'-X$. We note that $W$ contains at least one vertex $u_i$.
Since $W$ is a connected component, any edge in graph $G$ one end of which lies in $W$ and the other end outside $W$ ends in the set $X+u$.
Since the graph $G-X$ is connected, there exists an edge connecting $u$ and $W$. So some $u_i\in W$. It follows that the graph $F-X$ is connected.
I will be glad to answer any questions you may have about this reasoning.
