Let $(H, \langle\cdot,\cdot\rangle)$ be a Hilbert space and $\mathcal{M}\subset H$ some subset with $V:=\mathrm{span}(\mathcal{M})$.
Given a sequence $(x_k)$ in $H$, suppose that the functionals $\varphi_k:=\langle x_k, \cdot\rangle$ converge uniformly on $\mathcal{M}$.
Now if $V$ is not closed, can the limit of the operator sequence $(\varphi_k)$ still be uniquely extended to a bounded linear functional $\varphi : H\rightarrow \mathbb{R}$, i.e.: is there $x\in H$ such that $\varphi_k\rightarrow \langle x, \cdot\rangle$ uniformly [or at least pointwise] on $\mathcal{M}$?
(If $V$ is closed, the claim should hold by Banach-Steinhaus; yet in the given situation, the missing completeness of $V$ seems problematic.)
Let $\{e_n\}_{n=1}^\infty$ be an orthonormal basis in $\mathcal{H}$ and $\mathcal{M}=\{e_n\}_{n=1}^\infty.$ Consider $x_k=\sum_{n=1}^kn^{-1/2}e_n.$ Then for $l>k$ we get $$ |\varphi_l(e_n)-\varphi_k(e_n)|=|\langle e_n,x_l\rangle -\langle e_n,x_k\rangle| \\ =\begin{cases}n^{-1/2} & k< n\le l\\ 0 & {\rm otherwise} \end{cases}\quad <k^{-1/2}$$ Thus $\varphi_k$ satisfy the uniform Cauchy condition on $\mathcal{M}.$ However the sequence $\{x_k\}_{k=1}^\infty$ is unbounded. Therefore such $x$ does not exist.