Suppose $V$ is a vector space and $A \subset V$ such that $A$ is closed under affine linear combination i.e. $$\sum a_iv_i \in A, \text{ if } v_i \in A \text{ and } \sum a_i = 1 $$ then prove that $A = x_0 + W$ for some vector $x_0 \in V$ and some subspace $W$.
My approach:
Since both $x_0$ and $W$ are unknown, I try to create a linear map $T$ such that ker$(T) = U$. I do so by choosing a vector $v_1 \in A$ and putting $T(v_1) = c$. Now I choose vector $v_2$ and put $T(v_2)= c$. Now it is clear that all affine combination of $v_1$ and $v_2$ will go to $c$ as $$T(c_1v_1 + c_2v_2) = (c_1 + c_2)T(v_1) = T(v_1) = c .$$ I repeat this process until I cover $A$ and hence I will get my result.
Problem is : I don't know if I will be able to complete the process in finitie number of steps (and will it be an issue if I am not able to do that).
One clue is that for any vector $x_0$ and any subspace $W$, if $A = x_0 + W$ then $x_0 \in A$: this holds because $0 \in W$ and so $x_0 = x_0 + 0 \in x_0 + W = A$.
So all you have to do is to pick any $x_0 \in A$, then define $W = A - x_0$, and then use that definition together with the properties of $A$ to prove that $W$ is a subspace.
So first you need to prove that if $w \in W$ and $r \in \mathbb{R}$ then $rw \in W$. You know that $w + x_0 \in A$, and you know that $x_0 \in A$. Since $r + (-r+1)=1$ it follows that $$rw + x_0= r \cdot (w+x_0) + (-r+1) \cdot (x_0) \in A $$ and so $rw \in W$.
And next you need to prove that if $w_1 \in W$ and $w_2 \in W$ then $w_1 + w_2 \in W$. You know that $w_1 + x_0 \in A$ and $w_2 + x_0 \in A$. Since $1 + 1 - 1 = 1$ it follows that $$w_1 + w_2 + x_0 = 1 \cdot (w_1 + x_0) + 1 \cdot (w_2 + x_0) - 1 \cdot (x_0) \in A $$ and so $w_1+w_2 \in W$.