Let $V$ be a vector space over $F$ such that $\dim(V) = n$. prove that $V$ is isomorphic to $F^n$. lSo, here's what I'm trying to prove:
Let $V$ be a vector space over $F$. Prove that if $\dim(V) = n$, then $V$ is isomorphic to $F^n$. 

Proof Attempt:
We need to construct a bijective linear transformation $T: F^n \to V$. Let $(e_1,e_2,\ldots, e_n)$ be a basis of $F^n$ and $(v_1,v_2,\ldots,v_n)$ be a basis of $V$. So, we define the map as follows:
$$\forall u \in F^n: T(u) = \alpha_1 v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n$$
Let us prove that this is linear:
$$T(u_1+u_2) = T(\sum_{k=1}^{n} \alpha_k e_k + \sum_{k=1}^{n} \beta_k e_k)$$
$$\implies T(u_1+u_2) = T(\sum_{k=1}^{n} [(\alpha_k+\beta_k)e_k])$$
c
$$\implies T(u_1+u_2) = \sum_{k=1}^{n} [(\alpha_k+\beta_k)e_k]$$
$$\implies T(u_1+u_2) = \sum_{k=1}^{n} \alpha_k e_k + \sum_{k=1}^{n} \beta_k e_k = T(u_1) + T(u_2)$$
Let $\beta \in F$. Then:
$$T(\beta u) = T(\beta \cdot \sum_{k=1}^{n} \alpha_k e_k) = T(\sum_{k=1}^{n}[\beta \alpha_k e_k])$$
$$T(\beta u) = \sum_{k=1}^{n} (\beta \alpha_k v_k) = \beta \cdot \sum_{k=1}^{n} \alpha_k v_k = \beta \cdot T(u)$$
This proves linearity. Now, we have to show that this is bijective. To do this, we need to demonstrate that $T$ is injective and surjective. We prove injectivity first. Let $u_1,u_2 \in F^n$. Then:
$$T(u_1) = T(u_2)$$
$$\implies \sum_{k=1}^{n} \alpha_k v_k = \sum_{k=1}^{n} \beta_k v_k$$
$$\implies \sum_{k=1}^{n} [(\alpha_k - \beta_k)v_k] = 0$$
$$\implies \forall k \in \{1,2,\ldots,n\}: \alpha_k = \beta_k$$
where that last implication holds because of the linear independence of $(v_1,v_2,\ldots,v_n)$. Now, we prove surjectivity. Let $v \in V$. Then:
$$\exists \alpha_1,\ldots,\alpha_n \in F: v = \sum_{k=1}^{n} \alpha_k v_k$$
We note that $v_k = T(e_k)$, as a result of our definition. So:
$$v = \sum_{k=1}^{n} \alpha_k \cdot T(e_k) = \sum_{k=1}^{n} T(\alpha_k \cdot e_k)$$
$$v = T[\sum_{k=1}^{n} \alpha_k \cdot e_k]$$
Define $u = \sum_{k=1}^{n} \alpha_k \cdot e_k$. So, there exists a $u \in F^n$ so that $T(u) = v$. Since $v$ was arbitrary, we have proven that $T$ is surjective. That proves that $T$ is a bijective linear transformation or an isomorphism.
Does the proof above work? If it doesn't, why? How can I fix it?
 A: Your proof sounds good. Here I propose a brief version based on your solution.
You can also consider the linear transformation $T:V\rightarrow\textbf{F}^{n}$ given by $T(v_{j}) = e_{j}$, which exists and is unique. Thus it suffices to prove that $T$ is injective (according to the rank-nullity theorem).
Indeed, this is the case:
\begin{align*}
T(v) = T(w) & \Longrightarrow T(a_{1}v_{1} + a_{2}v_{2} + \ldots + a_{n}v_{n}) = T(b_{1}v_{1} + b_{2}v_{2} + \ldots + b_{n}v_{n})\\\\
& \Longrightarrow a_{1}T(v_{1}) + a_{2}T(v_{2}) + \ldots + a_{n}T(v_{n}) = b_{1}T(v_{1}) + b_{2}T(v_{2}) + \ldots + b_{n}T(v_{n})\\\\
& \Longrightarrow a_{1}e_{1} + a_{2}e_{2} + \ldots + a_{n}e_{n} = b_{1}e_{1} + b_{2}e_{2} + \ldots + b_{n}e_{n}\\\\
& \Longrightarrow (a_{1} - b_{1})e_{1} + (a_{2} - b_{2})e_{2} + \ldots + (a_{n} - b_{n})e_{n} = 0\\\\
& \Longrightarrow v = w
\end{align*}
Thus $T$ is injective.
BONUS
Based on the same procedure, you can prove that, given any pair of finite dimensional vector spaces $V$ and $W$ such that $\dim V = \dim W$, they are isomorphic.
