Why wouldn't this proof be correct for ℘(A ∪ B) = ℘(A) ∪ ℘(B) ↔ (A ⊂ B) ∨ (B ⊂ A) ∨ (A = B) Let U={1, 2, 3, ... 24, 25, 26, a, b, c, ... x, y, z}, A = {1, 2, 3, ... 24, 25, 26} and B = {a, b, c ... x, y, z}, then 
→ A ∪ B = U
 → U ⊆ A ∪ B    Definition of a union
 → U ∈ ℘(A ∪ B)     Definition of a Power Set 
However,  U ⊈ A ∧ U ⊈ B
→ U ∉ ℘(A) ∧ U ∉ ℘(B)      Definition of Union & Power Set      
 → ~(U ∈ ℘(A) ∨ U ∈ ℘(B))  DeMorgan's Law           
 → U ∉ ℘(A) ∪ ℘(B)     Definition of Union & Power Set  
 → ℘(A ∪ B) ⊈ ℘(A) ∪ ℘(B)  Definition of a subset            
∴ ℘(A ∪ B) ≠ ℘(A) ∪ ℘(B)
However, Let U, A and B be arbitrary sets, then 
→ U ⊆ A ∪ B     Definition of a union
 → U ∈ ℘(A ∪ B)     Definition of a Power Set  
Then Let (A ⊂ B) ∨ (B ⊂ A) ∨ (A = B)
→ U ⊆ A ∨ U ⊆ B (Definition of Union & Power Set)
→ U ⊆ A ∪ B (Definition of Union)
→ U ∈ ℘(A) ∪ ℘(B) (Definition of Union & Power Set) 
∴ ℘(A ∪ B) = ℘(A) ∪ ℘(B) ↔ (A ⊂ B) ∨ (B ⊂ A) ∨ (A = B)
 A: "Let U, A and B be arbitrary sets"
If $A, B$ and $U$ are arbitrary why are you introducing $U$.  If $U$ is arbitrary, then it isn't mentioned in the statement at all. 
" then 
→ A ∪ B = U "
If $A \cup B = U$ then $U$ is not an arbitrary set.  You need to have this as a premise and not as a result.
You should have said "Let $U = A \cup B$ then..."
"→ U ⊆ A ∪ B Definition of a union"
This isn't the definition of a union but basic property of sets.  $S \subseteq S$ for all sets so $U \subseteq U = A \cup B$.
" → U ∈ ℘(A ∪ B) Definition of a Power Set "  
This isn't so much the definition as a consequence of identity.  $S \subseteq S$ for all sets so $S \in P(S)$ so $U \in P(U) = P(A\cup B)$.
Now at this point it isn't clear whether you have intended to have completed half of a two directional proof or are just starting a single direction.
"Then Let (A ⊆ B) ∨ (B ⊆ A) ∨ (A = B)" 
Those are 3 conditionals you can not assume all of them at once.  $A\subseteq B$ OR $B\subseteq A$ OR $(A= B)$.
"→ U ∈ ℘(A) ∪ U ∈ ℘(B)"
This is either wrong or meaningless.
It's not really okay to write $R \in S \in T$ but, I suppose many people can use it as shorthand to mean $R \in S$ and $S \in T$.  If that is what you meant:
$U \not \in P(A) \cup U$ and $P(A) \cup U \not \in P(B)$.
$U \in P(A)$ only if $U \subset A$ but $U = A \cup B$ so $U \not \subset A$ unless $B \subset A=U$.  $U \not \in U$ as we are using set axioms that do not allow sets to be elements of themselves.
$P(A) \cup U \in P(B)$ means that $P(A) \subset B$ and $U \subset B$.  $U \not \subset B$ unless $A \subset B = U$.  $P(A) \subset B$ means $B$ includes all the subsets of $A$ including $A$ itself.  THis could be true but we have utterly no reason to think it would be.  If it were than $A \not \subset B = U$ as our set axioms won't allow that.
"→ U ⊆ A ∨ U ⊆ B (Definition of Union & Power Set)"
This just isn't true at all.  $U = (A \cup B) \not subset A$ (unless $B \subset A$ nor is $U = (A \cup B) \not \subset B$ (unless $A \subset B$).
"→ U ⊆ A ∪ B (Definition of Union)"
You already stated this and this is certainly NOT the definition of union.
"→ U ∈ ℘(A ∪ B) (Definition of Power Set) "
You already stated this.
"∴ ℘(A ∪ B) = ℘(A) ∪ ℘(B) ↔ (A ⊆ B) ∨ (B ⊆ A) ∨ (A = B)"
Why?  I don't see how any of this follows.  You've stated $A\cup B \in P(A\cup B)$ (which is obvious) and nothing about $P(A) \cup P(B)$. 
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My advise:
Don't introduce $U = A\cup B$.  It adds nothing to your proof and it lead you on some very false roads.
Note: if $A \subseteq B$ then $S \subset A \implies S \subset B$.  So $(A \subset B) \lor (B \subset A) \lor (A =B) \implies P(A\cup B)= P(A) \cup P(B)$ should be obvious.
Note: if neither $A \subseteq B$ nor $B\subseteq A$ then there are $a \in A; b \in B; a \not \in B; b \not \in A$. 
So the set $\{a,b\} \subset A\cup B$  so $\{a,b\} \in P(A\cup B)$.  But $\{a,b\} \not \subset A$ and $\{a,b\} \not \subset B$.  So $\{a,b\} \not \in P(A) \cup P(B)$.
