Convergence equivalence If $G_k(\mathbb{R}^m)=\{ W: W$ is subspace of $\mathbb{R}^m, \dim W=k \}$ and consider in $G_k(\mathbb{R}^m)$ one topology $\tau$ where $U\in \tau$ is open iff  the set $\widehat{U}=\lbrace v: v\in W\backslash \lbrace 0\rbrace, \mbox{for some}  \ W\in U \rbrace$ is open in $\mathbb{R}^m$
I'm testing the following:
If for every $k\in \mathbb{N}$ there is a basis $\lbrace u^k_1,\ldots,u^k_n\rbrace$ of $S_k$ such that $\Lambda=\lbrace \displaystyle{\lim_{ k \rightarrow +\infty}}u^k_1,\ldots,\displaystyle{\lim_{ k \rightarrow +\infty}}u^k_n\rbrace$ is a basis of $S$ then $\lbrace S_k\rbrace \subset G_n(\mathbb{R}^m)$ converges to $S\in G_n(\mathbb{R}^m)$.
My progress:
Let $U$ open of $S$ in $G_n(\mathbb{R}^m)$. Given that $\widehat{U}$ is open in $\mathbb{R}^m$ such that $\Lambda \subset \widehat{U}$, there $k_0 \in \mathbb{N}$ such that $u^k_j \in \widehat{U}$ for all $k\geq k_0$ and all $1\leq j\leq n$. There $W_j\in U$ such that $u^k_j\in W_j$. But as showing $W_j=S_k$ for all  $1\leq j\leq n$?
Note: The other direction of the statement is also true but I only need the above.
This is an example of open

 A: $\DeclareMathOperator{\vecspan}{span}$The statement you wanted to prove is actually false, if I'm not mistaken. Here's a counterexample. Define $A \subset G_2(\mathbb R^3)$ by
$$
A = \left\{\vecspan\bigl\{(1,0,0),(0,1,t)\bigr\} : -1 < t < 1 \right\} \cup \left\{\vecspan\bigl\{(0,1,0),(1,0,t)\bigr\} : -1 < t < 1 \right\}.
$$
This is an open neighborhood of the element $S = \left\{ (x,y,z) : z = 0 \right\} \in A$ because
$\widehat A = \left\{ (x,y,z) : \lvert z \rvert < \max(\lvert x \rvert, \lvert y \rvert) \right\}$
is open.
Now define $\Lambda_k = \bigl( (1,0,1/k), (0,1,1/k) \bigr)$ and $S_k = \vecspan \Lambda_k$ for $k \in \mathbb Z_{>0}$. Note that $S = \vecspan(\lim_{k \to \infty} \Lambda_k)$.
However, we also have $S_k \notin A$ for all $k$ (a consequence being that $S_k$ does not converge to $S$). To see this, first note that $(1,0,1/k) \notin \vecspan \left\{ (1,0,0), (0,1,t) \right\}$ for $t \neq 0$; the reason for this is that when a point on that plane has nonzero $z$-coordinate, it must also have nonzero $y$-coordinate. Likewise, $(0,1,1/k) \notin \vecspan \left\{ (0,1,0), (1,0,t) \right\}$ for $t \neq 0$. And of course, neither $(1,0,1/k)$ nor $(0,1,1/k)$ lie on $S$. We have thus determined that each element of $A$ fails to contain at least one point of $S_k \supset \left\{ (1,0,1/k), (0,1,1/k) \right\}$. The desired conclusion follows.

Appendix. Judging by the critical response, some clarifications might be in order. While $k>2 \implies S_k \setminus \{0\} \subset \widehat A$, it nonetheless remains the case that $S_k \notin A$ for all $k \in \mathbb Z_{>0}$. The set $A$ is the union of the two families of planes specified at the top of the answer, nothing more, nothing less. In general, the implication $U \in A \implies U \setminus \{0\} \subset \widehat A$ is true, but the converse isn't.
There might also be confusion stemming from trying to use intuition from the canonical manifold topology on the Grassmannians. The $\tau$ topology from the question is strictly finer than the manifold topology. Thus open neighborhoods in $\tau$ do not necessarily contain every subspace that is sufficiently "close" in the usual manifold sense; while $(S_k)$ converges to $S$ in the manifold topology, the same sequence fails to converge in the $\tau$ topology.
