Can someone please check, and if necessary, correct my prrof? Thanks
$$ E = F1 \oplus F2 $$
Since B1 is a basis for F1 and B2 is a basis for F2, prove that B1 U B2 is a basis for E:
Any point of a vectorial space can be described as a linear combinatiom of its basis: $$ a \in F1\\ a = \alpha B1 : \alpha \in R $$ Same applies to F2: $$ b \in F2\\ b = \gamma B2 : \gamma \in R $$ Therefore, for any point that it's in the sum of F1 and F2, we have: $$ a+b \in E\\ a+b= \alpha B1 + \gamma B2 $$ By that we can assume that B1 U B2 is a basis for E.
-- UPDATED ANSWER --
I was considering B1 and B2 basis of 1 vector... But it's actually the set of all L.I vectors. Because of that I'll remake my answer:
Any point of a vectorial space can be described as a linear combination of the vectors of its basis: $$ a \in F1\\ a = \sum_{i=1}^{dim(F1)} \gamma_{i}\cdot V_{i} : V_{i} \in B1, \gamma_{i} \in R \\ $$ Same applies to F2: $$ b \in F2\\ b = \sum_{i=1}^{dim(F2)} \alpha_{i}\cdot V_{i} : V_{i} \in B2, \alpha_{i} \in R \\ $$ Therefore, for any point that it's in the sum of F1 and F2, we have: $$ a+b \in E $$ Since a and b are described by linear combinations of their basis B1 and B2 respectively, we can assume that $ B1 \cup B2 $ is a basis for E.
Your answer is on the way to halfway correct. Your notation $\alpha B1$ is not standard. If it means what I think it means - "a linear combination of elements of $B1$" - then you have proved everything in the sum is a linear combination of elements from the two bases.
It seems from your comment on another answer
that you are a little confused. A "linear combination of basis vectors" is the same as "in the span of the basis" but is not the same as "a scalar times the basis".
I suggest that you rewrite the proof from scratch without the $\alpha B1$ notation, saying at each step just what you know. For example,
You have to prove that representation is unique to finish. The parenthetical phrase above will help you.