My textbook on functional analysis says as follows.(The book is written in Japanese, ISBN: 978-4-946552-18-2)
Let $\{X_n\}$ be a sequence of locally convex spaces over $K (= \mathbb{R}$ or $\mathbb{C}$).
Suppose that $X_n$ is a subvector space of $X_{n+1}$, and the topology of $X_n$ is identical with the relative topology of $X_{n+1}$ for all $n \in \mathbb{N}$.
Then $\displaystyle X := \bigcup_{n \in \mathbb{N}}X_n$ is a vector space by
addition $f_{a}$: $X \times X \to X$
scalar product $f_{s}$: $K \times X \to X$,
where since for any $x, y \in X$, there exists $n_0 \in \mathbb{N}$ s.t. $x, y \in X_{n_0}$, we define $f_a(x, y) :=$ "the addition of $x$ and $y$ in $X_{n_0}$"(well-defined), and $f_s$ is also defined in the same way.
Furthermore, $X$ is a locally convex space if we equip $X$ with the topology which is generated by the following basis of neighborhoods of $x_0 \in X$: \begin{equation} x_0 + \{U \subset X \mid \textrm{$U$ is absolutely convex,} \\ \textrm{ and $U \cap X_n$ is a neighborhood of the origin of $X_n$ for all $n \in \mathbb{N}$}\} \end{equation}
I don't know why $X$ is a locally convex space.
To show $X$ is a locally convex space, we have to show the continuity of $f_a$ and $f_s$,
but I can't do that. Could anyone help me? Thank you in advance.
Definition of a term:
$U \subset X$ is absolutely convex $\iff$ For $\lambda \in K$ satisfying $|\lambda| \leq 1$, $\lambda U \subset U$ and for $x, y \in U$ and $t \in [0, 1]$, $tx + (1-t)y \in U$
I was able to show that in almost the same way as that in this link, replacing the argument in $\mathbb{R}$ with that in $\mathbb{C}$. Though when you see the linked page, you will know, remarking "$U \cap X_n$ is a neighborhood of the origin of $X_n$ for all $n \in \mathbb{N}$" implies $U$ is absorbing, my claim also easily follows from the proposition proved in the link.