Let $C\in\mathbb{R}^{n}$ be a convex set with $\text{aff}(C)=W+t$. For $x\in\text{cl}(C)\backslash\text{reint}(C)$, show that there exists $v\in W$, $v\neq0$, such that $$\text{sup}_{x\in C}v^{T}z=v^{T}x$$ In here, $\text{aff}()$ is the affine hull, $W$ a linear subspace of $\mathbb{R}^{n}$, $t$ unknown, $\text{cl}()$ the closure and $\text{reint}()$ the relative interior.
In the answer (using the separator theorem) the following claim is made, which I do not understand: Since, $x\in\text{cl}(C)\subseteq\text{aff}(C)$, we see that $\text{aff}(C)=W+x$, but why?
I thought that since a linear subspace of $\mathbb{R}^{n}$, $W$, is infinite, we can choose an offset randomly in this linear subspace, $W$, and create the same affine space, $W+t$. Since, $x\in\text{cl}(C)\subseteq\text{aff}(C)$, we may set $t=x$ without altering the affine space $\text{aff}(C)=W+x$.
Is this reasoning correct?
Your reasoning is almost correct. You can choose any vector in aff($C$) as the offset, not in $W$. As $t,x$ are both elements of aff($C$), we have $t-x \in W$ and thus $$ \mathrm{aff}(C) = t + W = t - (t-x) + W = x + W.$$ You can see what is going on in this example ($n = 2$):