To prove A $\implies$ B or A $\implies$ C with a direct proof, I'm confused on what we can assume. Normally for $A \implies B$ we assume A then prove B is true.
For proving A $\implies$ B or A $\implies$ C, do we just assume A is true then show the following:
- B is true and C is false
- B is false and C is true
- B is true and C is true
Or are there other ways using a direct proof?


Is your $\Rightarrow $ used to mean logical implication, or is it used as the material implication? It is typically used to mean the former ... if you mean the latter then please use $\rightarrow$
This distinction is crucial!
It is true that $(A \rightarrow B) \lor (A \rightarrow C)$ is logically equivalent to $A \rightarrow (B \lor C)$, i.e. that $(A \rightarrow B) \lor (A \rightarrow C) \Leftrightarrow (A \rightarrow B) \lor (A \rightarrow C)$, and hence it is true that $A \rightarrow (B \lor C) \Rightarrow (A \rightarrow B) \lor (A \rightarrow C)$, and therefore proving that $A \rightarrow (B \lor C)$ will indeed prove that $(A \rightarrow B) \lor (A \rightarrow C)$
However, it is not true that $A \Rightarrow B \lor C$ is the same as $A \Rightarrow B$ or $A \Rightarrow C$: if it is true that $A \Rightarrow B$ or $A \rightarrow C$, then it will also be true that $A \Rightarrow B \lor C$, but the other way around does not hold.
Consider:
$B=P$, $C = \neg P$, and $A=Q$
Now, clearly you have $Q \Rightarrow P \lor \neg P$, and yet you have neither $Q \Rightarrow P$ nor $Q \Rightarrow \neg P$
So, showing that $A$ logically implies $B \lor C$ by showing that the assumption that $A$ is true means that at least one of $B$ and $C$ has to be true shows that $A \Rightarrow B \lor C$, but that does not show that either $A$ logically implies $B$ or that $A$ logically implies $C$, i.e. your method would not show that $A \Rightarrow B$ or $A \Rightarrow C$