Assume $E$ is Banach space and $u,v\in\mathcal{B}(E)$. Prove that if $u(E)\subset v(E)$, then there exists some constant $k\geq 0$ such that for all $x\in E$, there exists $y\in E$ such that $\|y\|\leq k\|x\|$ and $u(x) = v(y)$.
Since $u(E)\subset v(E)$, for any $x\in E$, $u(x)\in v(E)$. Thus there exists some $y\in E$ such that $u(x) = v(y)$. But how to show that $\|y\|/\|x\|$ is bounded from above?
You could mimic the proof of the open mapping theorem. But there is a nice trick to apply the open mapping theorem: The space $$X=\{(y,z)\in E\times E: u(y)=v(z)\}$$ is again a Banach space as a closed (by the continuity of $u$ and $v$) subspace of $E\times E$ endowed, e.g., with the norm $\|(y,z)\|=\max\{\|y\|,\|z\|\}$ (in category theory, $X$ is called a pullback of $u$ and $v$). The assumption implies that the first projection $$\pi:X\to E,\, (y,z)\mapsto y$$ is surjective and hence open, i.e., there is a constant $k\ge0$ such that, for every $x\in E$, there is $(y,z)\in X$ with $\|(y,z)\|\le k\|x\|$ and $\pi(y,z)=x$. This means $y=x$ and $z\in E$ satisfies $\|z\|\le\|(y,z)\|\le k\|x\|$ as well as $u(x)=u(y)=v(z)$.
I prefer capital letters for operators. Assume $U(E)\subset V(E).$ Let $\tilde{E}=E/\ker V.$ Then $\tilde{E}$ is a Banach space with respect to the norm $$\|[x]\|=\inf\{\|y\|\,:\, y\in E, \ y-x\in \ker V\}$$ Let $\tilde{V}:\tilde{E}\to E$ denote the quotient operator defined by $\tilde{V}[y]=Vy,$ where $[y]=y+\ker V.$ Then $\tilde{V}$ is injective and $\tilde{V}(\tilde{E})= V(E).$ It is well known that the operator $\tilde{V}$ is bounded and $\|\tilde{V}\|=\|V\|.$
For $x\in E$ there is $y\in E$ such that $Ux=Vy.$ Then $Ux=\tilde{V}[y].$ Define the operator $W:E\to \tilde{E}$ by $$Wx=[y]\quad {\rm iff}\quad Ux=\tilde{V}[y]=Vy$$ Since $\tilde{V}$ is injective, the operator $W$ is well defined and linear. We will prove that $W$ is bounded by showing that its graph is closed. To this end let $x_n\to 0$ and $[y_n]:=Wx_n\to [y].$ We have $\tilde{V}[y_n]\to \tilde{V}[y].$ On the other hand $Ux_n\to 0$ hence $$0=\lim_nUx_n=\lim_n\tilde{V}[y_n]=\tilde{V}[y]$$ Thus $[y]=0,$ as $\tilde{V}$ is injective.
By construction we have $$U=\tilde{V}W,\quad W:E\to \tilde{E},\ \tilde{V}:\tilde{E}\to E\quad (*)$$ Let $x\in E.$ Then $Ux=\tilde{V}(Wx).$ Denote $Wx=[y].$ For every $x\in E$ there is an element $y_x\in y+\ker V$ such that $$\|y_x\|\le 2\|[y]\|=2\|Wx\|\le 2\|W\|\,\|x\|$$ Moreover $$Ux=\tilde{V}(Wx)=\tilde{V}([y])=Vy=Vy_x$$
Remarks
$1.$ The representation $(*)$ may be useful for other purposes.
$2.$ If $E$ is a Hilbert space, we do not need to use quotient spaces, but deal with orthogonal complements and operators defined on the entire space.
