Number of final states in a minimal DFA
Yes, the number of final states in a DFA is at least the number of final states in its minimal DFA. Your proof is correct at some level of formality.



Intuitively, as you have noticed, any DFA for $L$ can be reduced to the minimal DFA for $L$ by merging states, which must merge any final states to a final state.  



Here is how to make your proof mathematically more formal. Let $L$ be a language over $\Sigma$ and $D=(Q,\Sigma,\delta,s,F)$ be a DFA that accepts $L$. Let $M$ be the set of Myhill-Nerode equivalence classes with respect to $L$. 

Define a map $m:Q\to M$ such that $m(q)=[w]$, where $w$ is any input that will drive $D$ to state $q$ and $[w]$ is the Myhill-Nerode equivalence class that contains $w$. We can verify that $m$ is well-defined, i.e., $[w_1]=[w_2]$ if both $w_1$ and $w_2$ drive $D$ to $q$.

We can verify or recall the following claims, thus completing this proof.


$m$ maps a final state in $D$ to a Myhill-Nerode equivalence class that contains a word in $L$.
all final states in the minimal DFA of $L$ are in one-to-one correspondence to all Myhill-Nerode equivalence classes that contain a word in $L$. 




Here is a simple related exercise.

Exercise. The number of non-final states in a DFA is at least the number of non-final states in its minimal DFA.