Certainly not: For any field $\mathbb{F}$, consider the basis $((1, 0), (0, 1))$ of $\mathbb{F}^2$. Neither $(1, 0)$ nor $(0, 1)$ spans the subspace spanned by $(1, 1)$.
Yep, your proof is absolutely correct, your "obvious" part, in my opinion, is perfectly valid and should be obvious to anyone with a basic understanding of what a set operation is and linear algebra.
One thing to note is that you can avoid the basis extension theorem (though it is perfectly valid) in favour of a mathematical logical argument. Since $U$ is a proper subspace of $V$, there is some $a \in V$ that is not in $U$ and $a \neq \vec{0}$ since the zero vector is in $U$.
Notice that the subspace spanned by $a$, call it $W$ has the property that $U \cap W = \vec{0}$, since if $\vec{a}$ is not in $U$, then $c\vec{a}, c \in \mathbb{R} \neq 0$ also cannot be in $U$, otherwise contradicting the fact that $U$ is a subspace closed under scalar multiplication...
Anytime you add in some vector $a$ that is not in $U$, no linear combination of it can be in $U$. Since if some linear combination is in $U$, but $a$ is not in $U$, this contradicts its subspace assumption.
You can extend this argument to keep adding some vectors not in $U$ and can conclude that indeed, there is some subspace $U'$ such that $U + U' = V$
Best Answer
Let $\mathcal F$ be the set of those subsets $J$ of $I$ such that its $\operatorname{span}\bigl(\{v_i\mid i\in I\}\bigr)\cap W=\{0\}$. You can deduce from Zorn's lemma that $\mathcal F$ has a maximal element $M$. Then $\{v_i\mid i\in M\}$ is a basis of a complement of $W$ in $V$.