Let $E=\bigcup_n E_n$ be a LF space: each of the step $E_n$ is a Fréchet space, $E_n\subset E_{n+1}$ with continuous inclusion and $E$ is equipped with the inductive limit locally convex topology. Let $F$ be another Fréchet space equipped with a continuous inclusion $F\hookrightarrow E$ that factorizes through a closed embedding $F\hookrightarrow E_n$ for $n$ big enough. Assume that $E$ is Hausdorff, does it imply that $F\hookrightarrow E$ is a closed embedding (i.e. the topology on $F$ is induced from the one on $E$) ? If not, is it true if we assume furthermore $E$ to be complete ?
Here is a concrete example where I am not even able to decide if the above holds or not: let $\mathcal{O}_C$ be the LF space of very slowly increasing functions on $\mathbb{R}$. It contains all the $C^\infty$ functions $f:\mathbb{R}\to \mathbb{C}$ satisfying the following condition: there exists some $n\geqslant 0$ such that for every integer $d\geqslant 0$ the function $x\mapsto \frac{f^{(d)}(x)}{(1+\lvert x\rvert)^n}$ is bounded. Denoting by $\mathcal{O}_{C,n}$ the subapces of functions verifying the previous condition for a particular $n$, $\mathcal{O}_{C,n}$ is naturally a Fréchet space and $\mathcal{O}_C=\bigcup_n \mathcal{O}_{C,n}$. Now let $C^\infty([-1,1])$ be the space of smooth function on $\mathbb{R}$ with support in $[-1,1]$. Then, for every $n\geqslant 0$, $C^\infty([-1,1])$ is obviously a closed subspace of $\mathcal{O}_{C,n}$ and even more: $C^\infty([-1,1])$ is a closed subspace of $\mathcal{O}_C$. But, does the topology on $\mathcal{O}_C$ induce the natural Fréchet topology on $C^\infty([-1,1])$ ?