Don't we all have a favorite letter?

Since E and G is not liked by Mickey,therefore his favorite alphabet must be the remaining one, which is J. Also G is not liked by Daisy.Therefore the remaining distinct alphabet E must be her favorite.

The following grid is a summary of the problem statement:

$\begin{array} {| c | c | c | c | c |} \text{Character \ Favorite Letter} & E & G & J & L \\ \hline \text{Mickey} & \times & \times & & \\ \hline \text{Minnie} & & & & \\ \hline \text{Donald} & & & & \checkmark \\ \hline \text{Daisy} & & \times & & \end{array}$

Because everyone has a distinct favorite letter, no one other than Donald likes the letter L:

$\begin{array} {| c | c | c | c | c |} \text{Character \ Favorite Letter} & E & G & J & L \\ \hline \text{Mickey} & \times & \times & & \times \\ \hline \text{Minnie} & & & & \times \\ \hline \text{Donald} & \times &\times & \times & \checkmark \\ \hline \text{Daisy} & & \times & & \times \end{array}$

Now, from this new grid, the only person whose favorite letter is G must be Minnie, and similarly, the only person whose favorite letter is J must be Mickey. Hence the following grid

$\begin{array} {| c | c | c | c | c |} \text{Character \ Favorite Letter} & E & G & J & L \\ \hline \text{Mickey} & \times & \times & \checkmark & \times \\ \hline \text{Minnie} & & \checkmark & \times & \times \\ \hline \text{Donald} & \times &\times & \times & \checkmark \\ \hline \text{Daisy} & & \times & \times & \times \end{array}$

We can now complete the grid.

$\begin{array} {| c | c | c | c | c |} \text{Character \ Favorite Letter} & E & G & J & L \\ \hline \text{Mickey} & \times & \times & \checkmark & \times \\ \hline \text{Minnie} & \times & \checkmark & \times & \times \\ \hline \text{Donald} & \times &\times & \times & \checkmark \\ \hline \text{Daisy} & \checkmark & \times & \times & \times \end{array}$

Hence, by interpreting the grid above, we know that Daisy's favorite letter is E.