There is another proof that gives an alternative description of super connected graphs in terms of spanning trees (hence perhaps of interest):

The following statements are equivalent for any $n$-vertex graph $G$:

(i) $G$ is super connected.

(ii) $G$ is connected and in every spanning tree $T$ of $G$ every two leaves of $T$ are adjacent in $G$.

(iii) $G$ is the complete graph $K_n$ or the cycle $C_n$.

(i) $\Rightarrow$ (ii): This is because $T-a-b$ is connected, whenever $a$ and $b$ are two leaves in $T$.

(ii) $\Rightarrow$ (iii): Let $T$ be a depth-first-search (spanning-)tree of $G$, rooted at any vertex $v_0$. As the leaves of any DFS tree are pairwise non-adjacent in $G$, $T$ must be a Hamiltonian path, and the endvertices of $T$ must be adjacent in $G$. Thus, $G$ has a Hamiltonian cycle $C$, say $C: v_0v_1\ldots v_{n-1}v_0$. Assuming $G\not=C$, $C$ has a chord, say $v_0v_i$ for some $i \ge 2$. Write $A=\{v_{1},\ldots, v_{i-1}\}$, $B=\{v_{i+1},\ldots, v_{n-1}\}$. Then, first, every vertex in $A$ is adjacent to every vertex in $B$ because any $a\in A$ and any $b\in B$ are leaves in the spanning tree $C+v_0v_i-aa'-bb'$ of $G$, where $a'$, resp., $b'$ is a neighbor of $a$, resp., $b$ on $C$. Next, every two vertices $v_s, v_t\in A\cup\{v_0,v_i\}$, $s\le t-2$, are adjacent because, by the fact above, we have a chord $v_pv_q$ for any $v_p\in A$, $s<p<t$, and any $v_q\in B$, and hence, as above, $v_s$ and $v_t$ are two leaves of certain spanning tree of $G$. Similarly, every two vertices in $B\cup\{v_0,v_i\}$ are adjacent. It follows that $G=K_n$.

(iii) $\Rightarrow$ (i) is obvious.