Problem: Let $B_1$ and $B_2$ be two Banach spaces and take $T\in\mathcal B(B_1, B_2)$ to a bounded linear operator between the two that is also bijective. Prove that $B_1 \setminus \text{Ker}(T)\cong \text{Range}(T)$ if and only if $\text{Range}(T)$ is closed.
You are allowed to assume that the restriction of $T$ to $B_1 \setminus \text{Ker}(T)$ denoted by $$T_{B_1\setminus\text{Ker}(T)}: B_1\setminus \text{Ker}(T) \to \text{Range}(T)$$ is an isomorphism whereby $$T_{B_1\setminus\text{Ker}(T)} \in \mathcal B (B_1\setminus \text{Ker}(T), \text{Range}(T))$$ $$T_{B_1\setminus\text{Ker}(T)} \in \mathcal B (\text{Range}(T), B_1\setminus \text{Ker}(T))$$
Confusion: I am mislead by the specification that $\text{Range}(T)$ be closed. In the following argument, I assume that $T$ itself is an injective operator so that $\text{Ker}(T) = \{0\}$ enabling me to write $T_{B_1\setminus\text{Ker}(T)} = T$. I prove that $T$ is a homeomorphism and that $B_1\setminus\text{Ker}(T)$ is not closed, which appears to produce the opposite result as that established in the question.
$B_1 \setminus \text{Ker}(T)\ncong \text{Range}(T)$ if $\text{Range}(T)$ is closed.
Where is my argument mistaken?
Lemma ($T$ is a homeomorphism): There exists $C_1, C_2 \in \mathbb R$ satisfying the following
$$||Tv||_{B_2}\leq C_1 ||v||_{B_1}$$ $$||T^{-1}v||_{B_1}\leq C_2 ||v||_{B_2}$$
As $B_1$ and $B_2$ are both metric spaces, the $\epsilon\delta$-definition of continuity will suffice, and they are both sequential spaces, which is to say that they are closed if and only if they are sequentially closed. For any $\epsilon > 0$, choosing $\delta_1 < \epsilon C_1^{-1}$ and $\delta_2 < \epsilon C_2^{-1}$ will prove that $T$ and $T^{-1}$ are continuous maps. $T$ is a homeomorphism and consequently open, closed, and continuous.
Lemma ($B_1\setminus \text{Ker}(T)$ is not closed): Consider $u \in B_1 \setminus \text{Ker}(T)$ and define $u_n = u/n$ ensuring that for any $n\in\mathbb N$,
$$||u_n|| = \frac{1}{n} ||u|| > 0$$ and $$||Tu_n|| = \frac{1}{n} ||Tu|| > 0$$
The Archimedean principle establishes the existence of an $n\in\mathbb N$ for all $\epsilon > 0$ obeying $$0 \leq ||Tu_n|| = \frac{1}{n}||Tu|| < \epsilon$$ The limit $$\lim_{n\to\infty} ||Tu_n|| = 0$$ implies $$\lim_{n\to\infty} Tu_n = 0$$ by definiteness of the norm and continuity of $T$ concludes $$T(\lim_{n\to\infty} u_n) = 0$$
Finally, $B_1\setminus \text{Ker}(T)$ is not sequentially closed, and as $B_1$ is sequential $B_1\setminus \text{Ker}(T)$ cannot be closed.