Let $X$ be a vector space. Suppose that $\{X_n\}_{n=1}^\infty$ is a sequence of vector subspaces such that
$X_n \subseteq X_{n+1}$ for all $n$,
Each $X_n$ is a locally convex topological vector space, and the topology on $X_n$ coincides exactly with the subspace topology that it inherits for $X_{n+1}$.
Then we can definie the so-called inductive limit topology on $X$, which is the strongest locally convex, topological vector space topology for which the injections $X_n \to X$ are all continuous.
I would like to prove that if $X_n$ is complete for all $n$, then $X$ is complete.
This fact is stated in Reed and Simon's Methods of Modern Mathematical Physics, Vol.1. Reed and Simon give a reference for the proof: A. and W. Robertson's Topological Vector Spaces. Unfortunately, this text is not available at my University library.
Here's what I have so far. Suppose that $\{x_n\}_{n=1}^\infty$ is Cauchy in $X$. First of all, consider the case when there exists some index $N$ so that $X_N \cap \{x_n\}$ is infinite. Then we can choose some subsequence $\{x_{n_k}\}_{k=1}^\infty$ so that$\{x_{n_k}\} \subseteq X_N$. Next, Let $U$ be an arbitrary balanced, convex, basis-neighborhood of $0$ in $X_N$. Using this lemma, it is possible to lift $U$ to a balanced, convex neighborhood $\tilde{U}$ of $0$ in $X$, with the property that $X_N \cap\tilde{U} = U$. Because $\{x_n\}$ is Cauchy in $X$, there is an $M \in \mathbb{N}$ so that
$$n,m \ge M \implies x_n -x_m \in \tilde{U}.$$
This in turn implies that
$$k,j \ge M \implies x_{n_k} -x_{n_j} \in \tilde{U} \cap X_N = U.$$
Therefore $\{x_{n_k}\}$ is Cauchy in $X_N$. By the completeness of $X_N$, there exists some $x \in X_n$ so that $x_{n_k} \to x$. To finish this case, we notice that:
$$x_n - x = (x_n - x_{n_k}) + (x_{n_k} - x)$$
So if $n$ and $k$ are made large enough, $x_n - x$ will reside inside any neighborhood $U$ of $0$ (to make this last step totally clear, we do need to recall the fact that, within any open any $V$ containing zero in $X$, we can find an open neighborhood $\tilde{V}$ of $0$ so that $\tilde{V} +\tilde{V} \subseteq V$).
I'm not sure how to complete the proof in the case where $\{x_n\} \cap X_N$ is finite for all $N$. In that case, I'm not able to extract the subsequence $x_{n_k}$, which was the key to the proof above.
Hints or solutions are greatly appreciated.
We will show that the second case cannot happen. If there are only finitely many $x_i$ in each $X_k$, we find subsequences $(y_i)$ of $(x_n)$ and $(Y_i)$ of $(X_n)$ (say $y_i = y_{n_i}$, and $Y_i = X_{m_i}$), such that $y_i \in Y_{i+1}\setminus Y_i$. As $Y_i$ is complete, it is closed in $X$, so by Hahn-Banach there is an $x_i^* \in X^*$ such that $x_i^*|_{Y_i} = 0$ and $x_i^*(y_i) = i - \sum_{k=1}^{i-1} x_k^*(y_i)$. Let $x^* := \sum_{i=1}^\infty x_i^*$. Then, on each $Y_i$, $x^*$ is a finite sum of continuous linear functionals, so $x^*|_{Y_i}$ is continuous. As $X$ carries the direct limit topology, $x^*$ is continuous on $X$. A continuous linear map on locally convex spaces maps Cauchy sequences to Cauchy sequences, but $x^*(y_i) = i$. Contradiction.
So, $X$ is sequentially complete.