Given A Real Vector Space, Any two choices of Basis Gives A Same Topology Here is the question, and I'm very confused and having no idea where to start it.

Let $V$ be an $n$-dimensional real vector space. If $\beta = \{v_1,\ldots,v_n\}$ is a basis of $V$, let $f_\beta : \mathbb{R}^n \to V$ be a linear map defined by
$$f_\beta (x_1,\ldots,x_n) = \sum_{i=1}^n x_iv_i$$
(Note: $f_\beta$ is a vector space isomorphism with inverse $$(f_\beta)^{-1}(v) = [v]_\beta$$ Also, $f$ is bijective, and using the theorem that  there is a unique topology $\mathcal{T}_\beta$ on $V$ such that $f_\beta$ is a homeomorphism from $(\mathbb{R}^n, \mathcal{T})$ to $(V, \mathcal{T}_\beta)$ where $\mathcal{T}$ is standard topology on $\mathbb{R}^n)$
(a) Show first that  if $\lambda$ and $\beta$ are bases of $V$, $\mathcal{T}_{\beta} = \mathcal{T}_{\lambda}$.
(b) Using (a), define standard topology $\mathcal{T}_V$ on $V$ by $\mathcal{T}_V := \mathcal{T}_{\beta}$ where $\beta$ is any basis of $V$.Let $V_1, V_2$ be finite-dimensional real vector spaces with respective standard topology $\mathcal{T}_{V_1}$ and $\mathcal{T}_{V_2}$. If $T: V_1 \to V_2$ is a linear map, Show that $T$ is continuous function from $(V_1, \mathcal{T}_{V_1})$ to $(V_2, \mathcal{T}_{V_2})$

 A: Can you prove the following fact?  
Theorem: Suppose that $\phi: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is a linear transformation.  Then $\phi$ is continuous.
If you can prove that, it follows that $\phi: \mathbb{R}^n \rightarrow \mathbb{R}^n$ being an isomorphism of vector spaces implies that $\phi$ is a homeomorphism of topological spaces (since then the inverse of $\phi$ is also a linear transformation, hence continuous).  I'll show you how to do (a) using this theorem.
Now, $f_{\beta}: \mathbb{R}^n \rightarrow V$ is a vector space isomorphism, and it is a homeomorphism when $V$ is taken in the $T_{\beta}$-topology, right?  Similarly $f_{\lambda}^{-1}: V \rightarrow \mathbb{R}^n$ is a vector space isomorphism and a homeomorphism when $V$ is taken in the $T_{\lambda}$-topology.
Let $\phi: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be the composition $f_{\lambda}^{-1} \circ f_{\beta}$.  Then $\phi$ is a vector space isomorphism (as a composition of isomorphisms), hence by the Theorem it must be a homeomorphism.
Let $E$ be an open set in $V$ in the $T_{\lambda}$-topology.  We want to show that $E$ is also open in the $T_{\beta}$-topology.  By definition, $E = f_{\lambda}(W)$, where $W$ is open in $\mathbb{R}^n$.  Since $\phi$ is a continuous, $\phi^{-1}(W)$ is open in $\mathbb{R}^n$, hence $f_{\beta}(\phi^{-1}(W))$ is open in $V$ in the $T_{\beta}$ topology.  But $$f_{\beta}(\phi^{-1}(W)) = f_{\lambda}(W) = E$$  This is one inclusion.  The converse is an identical argument, just use the fact that $\phi^{-1}$ is continuous.  
