Let $G$ be a simple connected graph with $diam(G)=2$ and $S\subset E(G)$ with l(G) edges where $l(G)$ is the edge connectivity of the graph such that $G-S$ is not connected. I want to show that at least one of the connected parts of $G-S$ is isomorphic to $K_1$ or to $K_a$ where a is the minimum degree of the vertices of the graph G.
Let $k$ be the edge-connectivity of $G$, so $|S|=k$, and $\delta(G)\geq k$. $S$ is a minimal disconnecting set of edges, so $G-S$ has exactly two components(!), call them $X$ and $Y$. Let $A$ be the vertices of $X$ that are endpoints of an edge of $S$. Let $B$ be the vertices of $Y$ that are endpoints of an edge of $S$. We claim that $X-A$ and $Y-B$ cannot both be nonempty: this would exhibit vertices at distance 3(!).
So we may assume that $X=A$. Let $|X|=s$, then $s\leq k$. The degreesum (in $G$) of the vertices of $S$ is at least $ks$. Exactly $k$ edges are leaving $S$, so the degreesum in $S$ is at least $k(s-1)$.
On the other hand the degreesum in $S$ can be at most $s(s-1)$(!). So we either have $s=1$ (in which case $X=K_1$) or $s=k$ (in which case $X=K_k$ and $\delta(G)=k$(!)).
Depending on your knowledge you may want to add a bit more evidence to the places marked with (!).