Let $V$ be a vector space on the field $K$ (on this case $\mathbb{R}$) and let $U$ and $W$ be subspaces of $V$. We say $V$ is the direct sum of $U$ and $W$, written as $V = U \bigoplus W$, if, given $v \in V$ there exist unique $u \in U $ and $w \in W$ such that $v = u + w$. We define $$U + W = \{v \in V; \text{there exist $u \in U$ and $w \in W$ such that $v=u+w$}\}.$$
Prove that $V = U \bigoplus W \iff V = U+W, U \bigcap W = \{0\}$.
I want to check whether or not my work is correct:
Proof: $U \bigcap W$ has to containt at least $0$ by definition. Assume it contains at least one element other than $0$, that is, $U \bigcap W = \{0, x_1 \}$. Then $2x_1 = x_1 + x_1 = 2x_1 + 0$, so $2x_1 \notin U \bigoplus W$, since that sum is not unique. So we've proven that if $U \bigcap W \neq\{0\}$, then $V = U \bigoplus W \neq U + W$, since there is at least one element of the latter that is not in the former. Now assume $U \bigcap W = \{0\}$ and take $x \in U+W$. Now suppose $U + W \neq U \bigoplus W$, that is, there exist different $a, c\in U$, $b, d\in W$, such that $x = a+b= c+d $. Then $a-c=d-b$, a contradiction. So now we've proven that, if, $V = U +W, U \bigcap W = \{0\}$, then $x \in U + W \implies x \in U \bigoplus W$. The other inclusion necessary is trivial, so we're done.
Is there anything wrong here? Did I miss anything? If it's all correct, can the writing or something else be improved? I'd appreciate any help.
Your proof seems to misinterpret the notation $U \oplus W$. This is a specific notation that is only used when $U + W$ is a direct sum. So it doesn't make sense to talk about whether a vector $u$ is in $U + W$ but not in $U \oplus W$. Either $U + W = U \oplus W$, or else it does not make sense to talk about $U \oplus W$.
Overall your proof seems to be correct, other than this misuse of notation. (i.e. instead of saying "$2x_{1} \not\in U \oplus W$" or $U+W \neq U \oplus W$" you would just say that "$U + W$ is not a direct sum".)