Any affine subspace of $\mathbb{R}^d$ is closed, in particular the intersection of all subspaces containing $X$. Therefore $\text{Aff}(X)$ is a closed subset of $\mathbb{R}^d$ containing $X$, hence also $\text{Closure}(X)$. Because $\text{Aff}(X)$ is the smallest affine subspace containing $X$, it is also the smallest affine subspace containing Closure($X$).
Edit: Let us prove that every affine subspace $A$ of $\mathbb{R}^d$ is closed. Pick $a \in A$ and set $W = -a +A$. It is clear that $A$ is closed if and only if $W$ is closed. Next, we wishes to write $W$ as the solution set to some finite set of equations, from which we can conclude that it is closed. For this, choose a basis $\{w_1,\ldots,w_n\}$ of $W$. Next, write
$$ w_i = \sum_{k=1}^d w_i(k) e_k, $$
and
$$ x = \sum_{k=1}^d x_k e_k, $$
where $e_1,\ldots,e_d$ is the standard basis of $\mathbb{R}^d$.
Then an element $x \in \mathbb{R}^d$ satisfies $x \in W$ if and only if there exist $\lambda_1,\ldots, \lambda_n \in \mathbb{R}$ such that
$$ x= \lambda_1 w_1 + \ldots + \lambda_n w_n. $$
This is equivalent to
$$ \begin{pmatrix} x_1 \\ \vdots \\ x_d \end{pmatrix} = \begin{pmatrix} w_1(1) &\cdots & w_n(1) \\
\vdots &\ddots & \vdots \\
w_1(d) &\cdots & w_n(d)
\end{pmatrix} \begin{pmatrix} \lambda_1 \\ \vdots \\ \lambda_n \end{pmatrix}. $$
Because the rank of the matrix $\begin{pmatrix} w_1(1) &\cdots & w_n(1) \\
\vdots &\ddots & \vdots \\
w_1(d) &\cdots & w_n(d)
\end{pmatrix} $ is $n \leq d$, Gaussian elimination will let you write this as
$$A \begin{pmatrix} x_1 \\ \vdots \\ x_d
\end{pmatrix} = \begin{pmatrix} \lambda_1 \\ \vdots \\ \lambda_n
\end{pmatrix} $$
for some matrix $A \in M^{n \times d}(\mathbb{R})$. Then $W = f^{-1}(\mathbb{R}^n)$, where $f$ is the continuous funcion $\mathbb{R}^d \rightarrow \mathbb{R}^n: x \mapsto Ax$.
