Faulty proof that $V=U_1 \oplus W$ and $V=U_2 \oplus W$ implies $U_1 = U_2$ The question is as follows:

Prove or give a counterexample: if $\ U_1, U_2, W$ are subspaces of $V$ such that
$V=U_1 \oplus W$ and $\ V = U_2 \oplus W$,
then $\ U_1 = U_2$.

I happily proved this but then found out that it is in fact incorrect! I posted a counterexample here. Could someone point out what is wrong with my 'proof'?

$V = U_1 \oplus W$ and $V=U_2 \oplus W$, so $V=U_1 + W$ and $V=U_2+W$, hence $U_1 + W = U_2 +W$.
Since $U_1 + W$ contains all elements $u_1 \in U_1$, we have that $u_1 \in U_2 + W$. Since $U_2 + W$ contains all elements of $U_2$ and $W$, we have that $u_1 \in U_2$ or that $u_1 \in W$.
If we assume, for contradiction, that $u_1 \notin U_2$ then $u_1 \in W$. But then, since $-u_1 \in W$, we are able to write $0 = u_1 + (-u_1)$. As $u_1 \neq 0$ (because $0 \in U_2$), we have written $0$ as a sum of two nonzero elements in $U_1$ and $W$ respectively. This contradicts that $V = U_1 \oplus W$.
Hence, $u_1 \in U_2$. This implies that $U_1 \subseteq U_2$. By the same line of reasoning it can be shown that $U_2 \subseteq U_1$, and thus $U_1 = U_2$.

$u\in U + W$ does not mean that $u\in U$ or $w\in W$; rather, it means that $u$ is the sum of something in $U$ and something in $W$.
