Let $V, H$ be real, separable Hilbert spaces and suppose that $V$ is dense in $H$. Since $V$ is separable, there exists a countable basis, namely $\{w_1, w_2,...\}$. Suppose we are given $u_0\in H$ and a sequence $(u_{n0})$ from $H$ such that $$u_{n0}\to u_o\qquad \mbox{in}\quad H.$$ Is it possible to construct $(u_{no})$ such that $u_{no}\in\mbox{span}\{w_1,...,w_n \}?$
I was thinking the following way. Suppose that $(v_n)$ is a sequence from $H$ and $v_n\to u_0$ in $H$. Since $V$ is dense in $H$ for every $v_n$, then there exists a sequence $(v_n^k)\subset V$ such that $$v_n^k\to v_n\qquad \mbox{in}\quad H.$$ As $V$ is separable Hilbert space, there exists $(\alpha_i^k)_{i\in \mathbb{N}}\in \mathbb{R}$ such that $$v_n^k=\sum_{i=1}^{\infty}\alpha_i^k w_i.$$ This means that $$\sum_{i=1}^{\infty}\alpha_i^k w_i\to v_n$$ as $k\to \infty$. Now I think we have to compare it with linear span in infinite-dimensional Hilbert spaces. If $\{w_1, w_2,...\}$ is a basis in $V$ it means that $$\mbox{span}\{w_1, w_2,...\}=\{\sum_{n=1}^{\infty}c_iw_i\,|,\ c_i\in \mathbb{R}\quad\mbox{and}\quad \sum_{n=1}^{\infty}c_iw_i \quad\mbox{converges unconditionally}\}.$$ The last one means that there are countable many nonzero elements? If so, we are done I suppose. Do you find it correct?