Is the Completeness of $X$ really necessary here?

Question

From an exercise in Kreyszig's Functional Analysis, it is stated that

Let $X$ be a Banach Space and $(x_n)$ be a sequence in $X$ such that the sequence $(f(x_n))$ is bounded $\forall f\in X'$, show that $(\|x_n\|)$ is bounded.

I attempted to prove it by letting $g_n\in X''$ be the canonical image of $x_n$ and invoked the Uniform Boundedness Theorem to show that the sequence $(\|g_n\|)$ is bounded. Since $\|g_n\|=\|x_n\|$, the sequence $(\|x_n\|)$ is bounded so the theorem is proved.

However, I noticed that nowhere in the proof did I once use the fact that $X$ is complete. The closest thing I used is the completeness of $X'$ but $X'$ is always a Banach Space regardless of $X$ so I am at lose.

So, is my proof correct? If it's not, then how could I use the completeness of $X$ to fix it? Thank you in advance.

Edit: It appears that $X$ being a Banach Space is really not necessary after all. Anyway, can anyone think of a way to prove it using the fact that $X$ is complete?

Answer 1

You are right. The result holds in every locally convex space (i.e., a vector space $X$ with a family $\mathcal P$ of semi-norms: $(X,\mathcal P)$ and $(X,\sigma(X,X'))$ have the same bounded sets.