Let $\{E_n, n\in \mathbb{N}\}$ be an increasing sequence of linear subspaces of a vector space $E$, i.e. $E_n \subset E_{n+1}$ for all $n\in \mathbb{N}$ such that $E = \bigcup_{n\in \mathbb{N}} E_n$. For each $n\in \mathbb{N}$, let $(E_n, τ_n)$ be a locally convex space such that the topology induced by $τ_{n+1}$ on $E_n$ coincide with $τ_n$. Equip $E$ with the (strict) locally convex inductive limit topology $τ$, i.e., the finest locally convex topology making all inclusions $E_n \subset E$ continuous.
Now let $D$ be a subset of $E$ such that for any $n\in \mathbb{N}$, $D_n=D\cap E_n$ is dense in $(E_n, τ_n)$. I am trying to prove that $D$ is dense in $(E, τ)$.
My attempt (please, check).
Let $U$ be a neighbourhood of the origin in $(E, τ)$. Then, by definition of $τ$, there exists convex, balanced, and absorbing neighbourhood $V$ of the origin in $(E, τ)$ such that $V \subseteq U$ and, for each $n\in \mathbb{N}$, $V_n=V \cap E_n$ is a neighbourhood of the origin in $(E_n, τ_n)$. Pick $x\in E$. There is $m\in \mathbb{N}$ such that $x\in E_m$. Since $V_m$ is a neighbourhood of the origin in $(E_m, τ_m)$ and $D_m$ is dense in $(E_m, τ_m)$, there is $y \in D_m$ such that $x-y \in V_m$. As $y \in x+V_m \subset x+V \subseteq x+U$, we are done.
Is it Ok? Thanks.
Your argument is correct, but I think it is worth presenting it in simpler settings:
We are given a topological space $X$ together with family of sub-spaces $X_i$ which cover the whole space. We are also given a subset $D\subset X$, such that, $\forall i$, $D\cap X_i$ is dense in $X_i$.
We need to show that $D$ is dense in $X$. i.e. that any non empty open subset $U$ in $X$ meets $D$. Clearly, there must be some $i$ for which $U\cap X_i$ is non-empty (since $X=\cup_i X_i$). For this $i$, $U\cap X_i$ must meet the dense set $D\cap X_i$ in $X_i$. So $U$ meets $D$. q.e.d.