Let me expand on Asaf's answer above. Now you already said that if you can prove that $T$ is injective then by the Rank - Nullity Theorem and the fact that $\dim V = \dim W$ it follows that $T$ is surjective too, and hence an isomorphism.
Definition of a basis: A collection of $n$ vectors $w_1, \ldots w_n$ is said to be a basis for a vector space $W$ if the following hold:
(1) Any $w \in W$ can be written as a linear combination of $w_1, \ldots w_n$. Viz. you can always find scalars $a_1, \ldots a_n \in F$ such that $w = a_1w_1 + \ldots a_nw_n.$
(2) The the vectors $w_1, \ldots w_n$ are linearly independent. This means to say that the only way of writing the zero vector $0$ as a linear combination of $w_1, \ldots w_n$ is as $0 = 0\cdot w_1 + 0\cdot w_2 + \ldots 0\cdot w_n.$
We say that the dimension of $W$ is the number of vectors in this collection, which in this case is $n$.
Suppose that there is a vector $x = a_1v_1 + \ldots a_nv_n$ in the kernel of $T$. Then this means that $T(x) = a_1T(v_1) + \ldots a_n T(v_n) = 0$ by definition of $x \in \ker T$ and using linearity.
However by assumption we know that $\mathcal{B} = \{T(v_1), \ldots T(v_n)\}$ is a basis for $W$. Therefore by (2) above we know that the only way to write $0$ as a linear combination of the $T(v_i)$ is when $a_1 = a_2 = \ldots = a_n = 0$.
Therefore our original vector $x = 0\cdot v_1 + 0\cdot v_2 + \ldots 0 \cdot v_n = 0$ so that $x = 0$. So we have proven that $\ker T \subseteq \{0\}$. Now for the reverse inclusion it is clear that $0 \in \ker T$ because $T$ is a linear transformation. Therefore $\{0\} \subseteq \ker T$ so that $\ker T = \{0\}$.
Hence you have proven that $T$ is injective.
Doing it the other way round: Suppose you did not know injectivity but wanted to prove surjectivity. By definition (1) above and the fact that the collection $\{T(v_1),\ldots T(v_n)\}$ is a basis for $W$, every vector $w \in W$ can be written as a linear combination of the $T(v_i)$. Viz. for all $w \in W$, there exist $a_1, \ldots a_n$ such that
$ w= a_1T(v_1 ) +\ldots + a_nT(v_n).$
But then $T$ is a linear transformation so that $a_1T(v_1) + \ldots a_nT(v_n) = T(a_1v_1 + \ldots a_nv_n).$ However $a_1v_1 + \ldots a_nv_n$ is just some vector lying in $V$. Furthermore our choice of the vector $w \in W$ a priori was just any vector in $W$. It follows that any vector $w\in W$ is the image of some vector in $V$. This is exactly saying that $T$ is surjective. Done.
Extra exercise: Prove that a linear map $T$ between finite dimensional vector spaces is injective iff its kernel is trivial.