Proof Check: The union of two linearly independent sets is linearly independent Let U and V be two subspaces of a vector space E such that $U ∩ V = \{0\}$. Prove that if $X ⊆ U$ and $Y ⊆ V$ are two linearly independent sets then so is $X \cup Y$.
So I believe I have a good start, but it may not hold (I'm still a beginner at conducting proofs..). Feel free to critique wherever necessary!
Origonal Proof:
$X$ and $Y$ are linearly independent. Thus, they do not contain the $0$ element. 
$X$ is a subset of $U$ and $Y$ is a subset of $V$, and $V \cap U = \{0\}$, so $X \cap Y = \{\emptyset\}$. (So, for any $ x \in X \cup Y$, $x \neq 0 $)
Suppose $X\cup Y$ is linearly dependent.
So, there exists $x \in X\cup Y$ and $y \in X\cup Y$ such that $x = y$ and they exist in the linear combination of other elements $\in X \cup Y$. 
i.e. $ x = (\lambda a + \mu b) =y $ for $a, b \in X \cup Y$ and scalers $\lambda$ and $\mu$ in the Field.
If $ (\lambda a + \mu b) = 0$, then we are done, for $x \neq 0 \neq y$.
Let $ (\lambda a + \mu b) \neq 0$. (Note: $a \neq 0 \neq b$, and $\lambda \neq 0 \neq \mu$) 
Then $a$ and $b$ are in a linearly independent set, so $X \cup Y$ is linearly independent.
Thank you for your help!
 A: Suppose $X \cup Y$ is not linearly independent. 
So, there exists $x_1, ... ,x_n \in X $ and $y_1, ..., y_n \in Y$ s.t. $\lambda_1 x_1 + ... + \lambda_n x_n + \alpha_1 y_1 +...+ \alpha_n y_n = 0$. 
$\lambda_1 x_1 + ... + \lambda_n x_n = -(\alpha_1 y_1 +...+ \alpha_n y_)$
$(\lambda_1 x_1 + ... + \lambda_n x_n) \in$ $span(X)$, which is contained in $U$, since U is a subspace.
$-(\alpha_1 y_1 +...+ \alpha_n y_) \in$ $span(Y)$ which is in $V$, similarly.
But $span(U \cap V)=\{0\}$, so the only solution to $\lambda_1 x_1 + ... + \lambda_n x_n = -(\alpha_1 y_1 +...+ \alpha_n y_)$ is if all of the scalers, $\lambda$ and $\alpha$ are $=0$
thus shows linear independence in $X \cup Y$
A: I think much of what you write is going around in circles. Rather than correct the errors, I'd suggest a different start. 
Suppose that union is not linearly independent. Then there is some nontrivial linear combination of the vectors in $X \cup Y$ that is $0$. Rewrite that so that it shows that a combination of the vectors in $X$ is equal to a combination of the vectors in $Y$, so is in the span of each, so is in $U \cap V$. 
What can you conclude from that? How does it contradict the linear independence of $X$ or $Y$?
A: Sorry; I can't follow your proof at all.
What you always want to do when proving results about linear (in)dependence is to recall how dependence is defined: that some linear combination of elements, not all coefficients zero, gives the zero vector. Suppose some such combination exists with elements of $X\cup Y$. The parts of the sum due to the respective elements of $X,\,Y$ are then some two vectors summing to zero, say $v$ and $-v$.
By thinking about which vector spaces $v$ provably lives in, can you prove $v=0$? By the linear independence of $X$, what does that tell you about some coefficients? Similarly with $Y$; what does that tell you about the other coefficients?
