We know the following classical statement: "Let $X$ a Banach space and $T:X\to X$ a bounded operator such that $\|T\|<1$. Then $I-T$ is invertible".
When we review the proof, it is easy to note how completeness of $X$ is required. But, do you know some example of a non Banach space $X$ such that there is a bounded operador $T$ with $\|T\|<1$ and $I-T$ is not invertible?
Consider the non-Banach space $c_{00}$ of all finitely-supported sequences equipped with the supremum norm. Let $S : c_{00} \to c_{00}$ be the unilateral shift. Consider $T = \frac12 S$. Then $\|T\| = \frac12 < 1$ but the inverse $(I-T)^{-1}$ is supposed to be given by $$(I-T)^{-1} = \sum_{n=0}^\infty T^n = \sum_{n=0}^\infty \frac1{2^n}S^n$$
However, this sum doesn't converge in the operator norm. Indeed, it doesn't even converge pointwise: for the canonical vector $e_1 \in c_{00}$ we have $S^ne_1 = e_{n+1}$ so it would be $$(I-T)^{-1}e_1 = \sum_{n=0}^\infty \frac1{2^n}S^ne_1 = \sum_{n=0}^\infty \frac1{2^n}e_{n+1} = \left(1, \frac12, \frac14, \ldots\right)\notin c_{00}$$
Therefore $(I-T)^{-1}$ doesn't exist so $I-T$ isn't invertible.
For a more formal proof that $I-T$ is not invertible, we'll show that $e_1$ is not in the image of $I-T$.
Indeed, assume that $\exists x = (x_n)_n \in c_{00}$ such that $(I-T)x = e_1$. Then
$$(1,0,0, \ldots) = e_1 = (I-T)x = \left(x_1, x_2 - \frac12 x_1, x_3-\frac12x_2, \ldots\right)$$
which implies $x_1 = 1$ and $x_{n+1} = \frac12 x_n, \forall n \in \mathbb{N}$. Induction gives $x_n = \frac1{2^{n-1}},\forall n \in \mathbb{N}$ or $$x = \left(1, \frac12, \frac14, \ldots\right)\notin c_{00}$$ which is a contradiction.