A group is called topological group if it satisfies three properties
1) G is a Hausdorff space in K (Here we want to prove that if $ A,B \in GL(2,\mathbb{R})$ then we can find two disjoint open sets $F,G$ in $\mathbb{R}^{4}$ such that $ A \in F,B \in G $
2)Multiplicative mapping is continuous (That is if $f :GL(2,\mathbb{R})$ x $ GL(2,\mathbb{R}) \longrightarrow GL(2,\mathbb{R})$ defined as $f((A,B))=AB$ then $f$ is continuous- Corresponding to open set $V \in GL(2,\mathbb{R})$ (in $\mathbb{R}^{4}$) , $f^{-1}(V)$ open in $GL(2,\mathbb{R})$ x $ GL(2,\mathbb{R})$ (in $\mathbb{R}^{4}$ x$\mathbb{R}^{4}$).
3)Inverse map is continuous ($g :GL(2,\mathbb{R}) \longrightarrow GL(2,\mathbb{R}) $ defined as $ g(A)=A^{-1}$.Corresponding to $V $ open $\in \mathbb{R}^{4}$ $f^{-1}(V)$ is open in $\mathbb{R}^{4}$.
What will be an open set in $\mathbb{R}^{4}$ ?
How can we prove that these open set contains a matrix $A$ ( surely 2x2 and non singular) ?
To verify the three properties:
1) $GL_2(\mathbb{R})$ is a subspace of the Hausdorff space $\mathbb{R}^4$ (assuming it's given the obvious topology).
2) For $A, B\in GL_2(\mathbb{R})$, each index $(AB)_{ij}$ is a polynomial in the $A_{pq}$ and $B_{pq}$ and thus continuous.
3) For $A\in GL_2(\mathbb{R})$ entries in the inverse matrix $A^{-1}$ are of the form $A_{ij} = p_{ij}/(\det A)$ for some polynomial $p_{ij}$ in the entries $A_{pq}$. The determinant $\det A$ is also a polynomial in the entries of $A$ and nonzero.
If I understand your question after the edit, you're asking about showing that any point in $\mathbb{R}^4$ has a neighborhood meeting $GL_2(\mathbb{R})$ (which isn't related to whether $GL_2(\mathbb{R})$ is a topological group). Since $\det A = A_{11} A_{22} - A_{12} A_{21}$ is a polynomial in the $A_{ij}$, any neighborhood of a point $A\in \mathbb{R}^4$ contains a point $A'$ with $\det A'\not = 0$; that is, $A'\in GL_2(\mathbb{R})$.