Is this sufficient for linear independence proofs?? I've been doing all of these proofs the same basically, I just want to make sure I'm doing them right, I didn't include all the details but I have the outlines of my proofs here.
1) U and W are subspace of V s.t. V = $U \oplus W$. Prove $u_1, \dots, u_m, w_1, \dots, w_m$ is a basis of V where the u's are a basis for U and the w's a basis for W.
I said that since the basis for U is lin ind in V then you can extend it to be a basis by adding the vectors from W. Since they are linearly independent as well, then you add every vector. This shows that $u_1, \dots, u_m, w_1, \dots, w_m$ is linearly independent. It obviously spans V because V = $U \oplus W$, there fore it is a basis
2) Prove or give counter example. $v_1, \dots, v_n$ and $v_1, \dots, v_n$ are linearly independent in V. Then:
$\lambda v_1, \dots, \lambda v_n$ with $\lambda \in F$ is linearly indp.
$\lambda (a_1 v_1 + \dots + a_n v_n) = 0$ (pulling out $\lambda$ from the span, we know the span is linearly indp). Therefore $\lambda 0 = 0$.
$v_1 + w_1, \dots, v_m + w_m$ is linearly indp.
Rearranging the span of this you get: $a_1 v_1 + \dots + a_m v_m + a_1 w_1 +\dots a_m + w_m = 0$ Split this into span(U) = $0$ and span(V) = $0$ implies all the coefficients must be zero.
 A: There are some glitches in your reasoning. In the first part you must use the hypothesis that $U\cap W=\{0\}$. In the second part, you're forgetting that $\lambda v=0$ when $\lambda=0$.
Part 1.
Let $\{u_1,u_2,\dots,u_m\}$ and $\{w_1,w_2,\dots,w_n\}$ be bases of $U$ and $W$ respectively. We want to prove that
$$\{u_1,u_2,\dots,u_m,w_1,w_2,\dots,w_n\}$$
is linearly independent. So, suppose
$$
a_1u_1+a_2u_2+\dots+a_mu_m+b_1w_1+b_2w_2+\dots+b_nw_n=0
$$
We must obviously use the fact that $U\cap W=\{0\}$ (which is part of the hypothesis $V=U\oplus W$), so we try to get a vector in $U\cap W$: set
$$
v=a_1u_1+a_2u_2+\dots+a_mu_m.
$$
By definition, $v\in U$. But we also have
$$
v=-(b_1w_1+b_2w_2+\dots+b_nw_n)
$$
so $v\in W$.
Then …
The part about $\{u_1,u_2,\dots,u_m,w_1,w_2,\dots,w_n\}$ being a spanning set is easy: if $v\in V$, then from $V=U\oplus W$ we know that $v=u+w$, with $u\in U$ and $w\in W$. Then we can write $u=\dots$ and $w=\dots$, so …
Part 2
If $\{v_1,v_2,\dots,v_n\}$ is linearly independent and $\lambda\in F$, we can consider
$$
a_1(\lambda v_1)+a_2(\lambda v_2)+\dots+a_n(\lambda v_n)=0
$$
from which
$$
\lambda(a_1v_1+a_2v_2+\dots+a_nv_n)=0
$$
that gives
$$
\lambda=0\qquad\text{or}\qquad a_1v_1+a_2v_2+\dots+a_nv_n=0
$$
Thus …
