It sounds like you're trying to show that, given a finite dimensional vector space $V$ over a field $F$, the space $S$ of linear transformations $T: V \to V$ is finite dimensional.
Here's a suggestion for how to proceed: First, show that $S$ is a vector space (if you haven't already). You can do this straight from the definition. (Show that if $T, U \in S$ and $\alpha, \beta \in F$ then $\alpha T + \beta U \in S$, which is to say that $(\alpha T + \beta U)(v) = \alpha T(v) + \beta U(v)$ is a linear transformation.)
Let $n = \dim V \in \mathbb{N}$. Fixing a basis allows us to identify $V$ with $F^n$ and $S$ with the set of all $n\times n$ matrices with entries in $F$. If you think about it, there should be a clear choice for a basis of this space $S$ over $F$ now.
If you're not seeing it, imagine taking the rows of an $n\times n$ matrix and laying them out in a line. Note you can do this for any $n\times n$ matrix, and so this allows you to think of $S$ as being essentially the same vector space as $F^{n^2}$. You now have an element of $F^{n^2}$. What's a basis for $F^{n^2}$? Relate this back to a basis for $S$. What's the order of the basis you found? It should be a function of $n$.
Ah. I just saw your edit. Note you're trying to find the dimension of $\dim S$, and so it's perfectly fine to set $\dim V = n$. In fact, you should expect that $\dim S$ is going to be a function of $\dim V$, and so this is the natural thing to do.
You don't need contradiction.
Suppose $\dim V<\dim W$; then
$$
\dim R(T)=\dim V-\dim N(T)<\dim W-\dim N(T)<\dim W
$$
so $\dim R(T)<\dim W$ and $T$ is not onto.
Suppose $\dim V>\dim W$; then
$$
\dim N(T)=\dim V-\dim R(T)>\dim W-\dim R(T)\ge0
$$
so $\dim N(T)>0$ and $T$ is not one-to-one.
What about your proofs? The fact that an onto linear map is also one-to-one is valid only if domain and codomain have the same dimension. So you can't use the fact that $T$ is one-to-one in the first attempt.
The second attempt is likewise affected by the wrong assumption that an one-to-one linear map is onto, which again is only valid if domain and codomain have the same dimension.
Best Answer
By Rank-Nullity Theorem,
$$\dim V = \dim N(T) + \dim R(T).$$
Thus, $$\dim R(T) = \dim V - \dim N(T) < \dim W.$$