Let $E$ be a Banach space, and $\operatorname{Isom}(E)$ the set of continuous bijections from $E$ to $E$. We define the set $U$ as: $$U = \{u \in \mathcal{L}(E) \mid u + \operatorname{id} \in \operatorname{Isom}(E) \} $$ Show that $U$ is open.
I attempted two methods, and in both of them I get stuck.
My first instinct was to try to show that the complementary is closed, but by looking at the complementary of the set. In this case we would have:
$$ \mathcal{L}(E) - U = \{u \in \mathcal{L}(E) \mid u + \operatorname{id} \not \in \operatorname{Isom}(E) \} .$$ But it doesn't seem to be easy to show that it's closed.
Secondly, I could show that for any element $x \in U$ there exists $r >0$ such that the open ball $B(x,r) \subset U$. But I am unable to come up with a potential candidate for $r$.
Any help would be highly appreciated.
As mentioned by Daniel Schepler in the comments, an important step is to show that if $\|u\|<1$, then $u+1$ is invertible. You should try to show this, but if you are unable almost every book on operator theory should contain a proof.
From here, suppose $u\in U$. Then $u+1$ is invertible (with bounded inverse by the open mapping theorem). If $v\in\mathcal L(E)$ and $\|u-v\|<\|(u+1)^{-1}\|^{-1}$, then $$\|1-(u+1)^{-1}(v+1)\|\leq\|(u+1)^{-1}\|\|u-v\|<1,$$ so $(u+1)^{-1}(v+1)$ is invertible, so $v+1$ is invertble and thus $v\in U$. Thus $B(u,\varepsilon_u)\subset U$ (where $\varepsilon_u=\|(u+1)^{-1}\|^{-1}$), and therefore $U$ is open.