Proof of uniform matrix being transversal matrix by selecting singletons

I am doing an assignment where I have to proof that every uniform matroid is a transversal matroid. My solution is a lot simpler than in the solutions manual, and normally this means that my solution is wrong. I just don't understand why. Is the following proof correct, and if not why?

First, the definitions:

Uniform matroid Given a set $E$ of size $n$ and a number $k \leq n$, the uniform matroid of rank $k$ has independent sets $J = \{I \subseteq E : |I| \leq k\}$

Partial Transversal Let $A$ be a family of subsets of a finite set $X$. A subset $I = \{ x_1,..,x_{|I|}\}$ of $X$ is called a partial transversal of $A$ if there are different sets $a_1,...,a_{|I|} \in A$ such that $x_i \in a_i$ for each $x_i$

Transversal matroid A matroid where the independent sets are the set of partial transversals

The question:

Proof that any uniform matroid is a transversal matroid.

My solution:

With set $E = \{1,..,n\}$

Take $A = \{\{i\} : i = 1,..,n\}$

Now for any independent set $I$ of the uniform matroid, each element of $I$ is in one of the sets of $A$.

I have a feeling i am completely misunderstanding the basics of a transversal matroid and missing something very obvious why my solution must be wrong, but i have no idea what it is.

Every independent set of the rank $k$ uniform matroid on $E$ is a partial transversal of $A$. However, there are more transversals of $A$, in particular $E$ itself. Thus, $A$ is a presentation for the uniform matroid of rank $|E|$.
The rank $k$ uniform matroid is presented by the set system $A_i = E$ for $i=1,\ldots,k$. There are other smaller presentations, but this is the largest possible presentation.