Let $X$ be an affine subspace and $a \in X$. We will prove that $X - a$ is a linear subspace, so let $x,y \in X$ and $\lambda, \mu \in \mathbb R$. Then
\[
\lambda(x-a) + \mu(y-a) = \lambda x + \mu y + (1 -\lambda - \mu)a - a
\]
Now $\lambda x +\mu y + (1-\lambda-\mu)a \in X$ as an affine combination of vectors from $X$, so
\[
\lambda(x-a) + \mu(y-a) = \lambda x + \mu y + (1 -\lambda - \mu)a - a \in X - a
\]
and $X-a$ is closed under linear combinations, hence a linear subspace.
Now suppose that $X-a$ is a linear subspace for some $a \in X$, then for $x_i \in X$ and $\lambda_i \in \mathbb R$ with $\sum_i \lambda_i = 1$ we have
\begin{align*}
\sum_i \lambda_i x_i &= \sum_i\lambda_i (x_i - a) + \sum_i\lambda_i a\\
&= \sum_i \lambda_i (x_i - a) + a\\
&\in X - a + a\\
&= X.
\end{align*}
Hence, $X$ is affine.
Note, that in arguing, we always supposed $X \ne \emptyset$, if $X$ is the empty affine subspace, the right hand side is trivially true.