Let $U$ be an arbitrary subspace of a vector space $V$ over a field $\Bbb F$ as follows. Proposition:
Let $V$ be a vector space over a field $\Bbb F$. A non-empty set $U \subseteq V$ called subspace of $V$ iff

*

*for all $u,w \in U$, then $u+w \in U$, and

*for all $k \in \Bbb F$ and $u \in U$, then $ku \in U$.

Below is the consequent of the proposition above, which I want to prove.
Let $U$ be an arbitrary subspace of a vector space $V$ over a field $\Bbb F$.
Then,

*

*$U$ is an additive subgroup of $V$, and

*$U$ contains $0*$, where $0*$ is a zero vector of $V$.

Attempt:
For $1$, let $U$ be an arbitrary subspace of a vector space $V$ over $\Bbb F$. Let $u,w \in U$. Denote $-w$ is the additive inverse of $w$ for all $w \in U$. Want to show: for all $u,w \in U, u - w \in U$.
This is follow by definition and by the proposition. Since $U$ is a subspace of $V$, then $U$ itself is a vector space over $\Bbb F$. So, for all $w \in U, -w \in U$ and hence $u+(-w) = u-w \in U$. Hence proved. $\Box$.
For $2$, by the proposition, there exists $0 \in \Bbb F$ such that for all $u \in U$, we obtain $0\cdot u = 0* \in U$. Thus, $U$ contains $0*$, the zero vector of $V$. Hence proved. $\Box$.
I know this is a very basic concept of vector space chapter. But, I only want to know that whether my answer above is correct. Thanks in advanced.
 A: Remember, to prove (i), we are simply proving that $U$ is in fact a subgroup of $V$ with respect to addition. This is equivalent to showing $U$ is closed with respect to addition and (additive) inverses, and that $U$ is non-empty. But by assumption $U$ is already non-empty, and $U$ is closed w.r.t. to addition. So we just have to show it is closed w.r.t. to inverses. But by the definition of what it means to be a subspace, $U$ is itself a vector space. Hence for any $\bar{u} \in U$, there must exist some additive inverse which we denote $-\bar{u}$. This simply follows from the axioms of what it means to be a vector space. We don't need to use the proposition at all (we just need to know that $U$ is a vector space, and you assume this anyway).
Addressing your actual proof of (i): your proof is correct, but it can be shortened. It's true that proving $\bar{u} - \bar{w} \in U$ shows it's a subgroup, but you just need to show it's closed wrt to negatives since by definition $U$ is non-empty and from the proposition it's obviously closed wrt to addition. I.e. just the final piece of your proof is what you need to actually prove the result.
For the proof of (ii), your proof is correct! Another way to see it is to  note that since $U$ is an additive subgroup of $V$. This means that $U$ must contain the identity element of $V$, since we have already proved that $U$ is closed w.r.t. inverses and addition. Therefore for any $\bar{u} \in U$, we have that $\bar{u} + (-\bar{u}) = 0* \in U$.
