How to prove:
An affine subspace $\mathbb{E}^n$ is $S=p+V$ for some $p\in\mathbb{E}^n$ and a vector space $V$ of $\mathbb{E}^n$.
I already tried showing $S-p=\{s - p \mid s \in S\}=V$ is subspace of $\mathbb{E}^n$. But it is hard to show that $V$ is closed under addition.
On
If $V=S-p$, and $x,y\in V$, then $x+p,y+p$ are in $S$, and thus:
$$m=\frac{1}{2}(x+p)+\frac{1}{2}(y+p)=\frac{1}{2}(x+y)+p\in S$$
But then $$2m+(1-2)p=x+y+2p-p=x+y+p\in S$$
And hence $x+y\in V$.
Similarly, if $x\in V$ and $\lambda$ is a scalar, then $x+p\in S$ and $p\in S$, so $$(1-\lambda)p + \lambda(x+p)=p+\lambda x\in S,$$ so $\lambda x\in V$.
Strong hint:
Assuming that an affine subspace is a subset $S$ closed under affine combinations, i.e., things of the form $(1-t)u + tv$, where $t \in \Bbb R$ and $u, v \in S$), then this isn't too tough.
Let $p$ be any element of the subspace $S$, and let $Q = \{ v - p \mid v \in S\}$. I'll show $Q$ is a vector subspace.
For $v \in Q$, let's show that $cv \in Q$ (where $c \in \Bbb R$), thus showing closure under scalar multiplication.
Well, $v \in Q$, so there's some $u \in S$ with $u - p = v$. Hence $$ cv = cu - cp $$ which is $$ cv = (cu - (c-1)p) - p\\ $$ so all I have to do is show that $h = cu - (c-1)p$ is in $S$, i.e., is an affine combination of two elements of $S$. Well, letting $t = 1-c$, we have $c = 1-t$, so $$ h = cu + (1-c) p = (1-t)u + tp $$ which is an element of $S$ because it's an affine combination of $u$ and $p$, which both are in $S$.
Now all you need to do is try to mimic that to show that $V$ is closed under addition as well.