Let us prove that every vector in U+V can be written as a linear combination of {u1,…,uk,v1,…,vl}.
U+V={u+v: u∈U, v∈V}
Since {u1,…,uk} is a basis of U, we can write u=α1u1+…+αkuk .
Since {v1,…,vl} is a basis of V, we can write v=β1v1+…+βlvl .
So, u+v=α1u1+…+αkuk+β1v1+…+βlvl .
Now we need to prove that {u1,…,uk,v1,…,vl} is linearly independent.
Suppose we have α1u1+…+αkuk+β1v1+…+βlvl=0 .
It can be written as ∑i=1kαiui=−∑j=1lβjvj .
We can see the left hand side is in U and the right hand side is in V . It means that both are in U∩V. Since U∩V={0} , it follows that ∑i=1kαiui=∑j=1lβjvj=0 .
Using the fact that {u1,…,uk} is a basis of U and {v1,…,vl} is a basis of V , we get that α1=…=αk=0 and β1=…=βl=0 .
So, {u1,…,uk,v1,…,vl} is linearly independent.
Now we can conclude that {u1,…,uk,v1,…,vl} is a basis for U+V .
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