# About a partial converse to the Banach-Steinhaus Theorem

I've been reading the GTM text Topics in Banach Space Theory by Albaic and Kalton. In the appendix, it states the following partial converse to the Banach-Steinhaus theorem:

Let $\{S_n\}$ be a sequence of operators from a Banach space $X$ into a normed linear space $Y$ such that $\sup_n\|S_n\|<\infty$. Then if $T:X\to Y$ is another operator, the subspace \begin{align*} M=\{x\in X:\|S_nx-Tx\|\to0\} \end{align*} is norm-closed in $X$.

I came up with the following proof of this result:

Put $K=\sup_n\|S_n\|$, let $x_0$ be a limit point of $M$, and let $\varepsilon>0$ be given. There exists $x\in M$ such that \begin{align*} \|x-x_0\|<\frac{\varepsilon}{2(K+\|T\|)}. \end{align*} For this $x$, there is some $N\in\mathbb{N}$ such that \begin{align*} \|S_nx-Tx\|<\frac{\varepsilon}{2} \end{align*} for $n\geq N$. For such $n$, we have \begin{align*} \|S_nx_0-Tx_0\|&\leq \|S_nx-Tx\|+\|S_n+T\|\cdot\|x_0-x\| \\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{align*} Thus $\|S_nx_0-Tx_0\|\to0$, and therefore $x_0\in M$. QED

However, in my proof the assumption that $X$ be a Banach space appears unnecessary. So my question is somewhat twofold: Is there something I overlooked in my proof? If not, is the assumption that $X$ is complete superfluous?

More generally, if $\{T, S_1, S_2, \ldots\}$ is a uniformly continuous family of maps from metric space $X$ to metric space $Y$, then $\{x \in X: \lim_{n \to \infty} S_n(x) = T(x)\}$ is closed. This does not require completeness of either metric space.
Your argument is fine, besides the typo where it should be $(\|S_n\|+\|T\|)$ instead of $\|S_n+T\|$.
And I don't think you need for $X$ to be complete.