Proof for $\mathbb{R}^{3}$ being the direct sum of two subspaces Given the two subspaces below
$$
U =  \left\{ (a,a,a)  \in R^{3}  : a \in \mathbb{R}  \right\} \\
V =  \left\{ (b,c,0)  \in R^{3}  : b,c \in \mathbb{R}  \right\} 
$$
Proof that  $R^{3} = U \oplus V$ and find basis of them:
1) $U \cap V = \left\{ 0  \right\}$, since subspace V has only vectors where the third coordinate is 0, and subspace U needs to have the same x, y and z coordinates . 
2) Since the intersection is $\left\{ 0  \right\}$ , given any $R^{3}$ vector, there's only one way to represent it using the subspaces given:
$$
Z = (x,y,z) \in R^{3} \\
Z = (a,a,a)+(b,c,0) : x = a+b, y = a+c
$$
3) Basis: 
For $U$: $(1,1,1)$
For $V$: $(1,0,0), (0,1,0)$
Is it correct? I'm new to linear algebra and math logic... 
Thanks!
 A: You basically have the right idea. For 2, it might be better to say something like this.
Let $(x,y,z)\in\mathbb{R}^{3}$. Then, $(z,z,z)\in U$, $(x-z,y-z,0)\in V$, and $(z,z,z)+(x-z,y-z,0)=(x,y,z)$. Thus, $\mathbb{R}^{3}=U+V$. Then, since you know that $U\cap V=\emptyset$, it follows that $\mathbb{R}^{3}=U\oplus V$.
Addendum. Here's why $U\cap V=\{(0,0,0)\}$. 
Suppose $(e,f,g)\in U\cap V$. Then, since it's in $U$, we know that $e=f=g$. Furthermore, since it's in $V$, it follows that $g=0$. Hence, $e=f=g=0$. Therefore, $U\cap V\subset\{(0,0,0)\}$. Clearly, $(0,0,0)\in U\cap V$, so we are done. 
A: Since we're beginning, you should probably justify (1) above, as the following: take $v \in U \cap V$. Then there are constants $a,b,c\in \Bbb R$ such that $v= (a,a,a) = (b,c,0)$, and follows that $a=b=c=0$, and so $v = 0$. Only now we've proven that $U \cap V = \{0\}$. Also, I feel that I should point that we're sort of abusing notation with this $\{0\}$ - it means the zero of the vector space $\Bbb R^3$, the point $(0,0,0)$.
For (2), the intersection being zero does not imply that that every vector in $\Bbb R^3$ writes as that convenient sum. Your answer to (3) is correct, and you can check that these vectors you found are linearly independent, and so they span $\Bbb R^3$, or alternatively, you can write $$(x,y,z) = a(1,1,1)+b(1,0,0)+c(0,0,1)$$and check that we can solve uniquely for $a,b$ and $c$ in terms of $x,y$ and $z$.
