This is an example similar to Bondy and Murty graph theory book. in that exercise $G$ is k-edge-connected and here it's 8 edge connected.
Show that if $G$ is 8-edge-connected, and if ${E}'$ is a sub set of $E$ edges of $G$, and $|{E}'|=8$ then $w(G-{E}')\leq 2$
$W$ is number of $G$ components.
Let $G=(V, E)$ and let $S$ be an arbitrary subset of $E$ containing exactly $7$ edges. As $G$ is $8$-edge-connected, the graph $G' = (V, E \setminus S)$ is connected. That means that $w(G') = 1$. Now choose any $e \in E \setminus S$ and remove it from $G'$. As $G'$ was connected, the graph $G'' = (V, E \setminus (S \cup \{e\}))$ is either connected or has exactly two components, because deleting a single edge can only increase the number of components by one. That means that $w(G'') \leq 2$.
As we chose both $S$ and $e$ arbitrarily, we have shown that removing any subset of edges of $E$ containing exactly $8$ edges can generate a graph with at most two components.