Minimum Least Squares Solution Using Pseudo Inverse (Derived from SVD) Is The Minimum Norm Solution - Extension from Vectors to Matrices 
Given $A \in \mathbb{R}^{m \times n}$, $B \in \mathbb{R}^{k \times \ell}$, and $C\in \mathbb{R}^{m \times \ell}$. Show that for $X \in \mathbb{R}^{n \times k}$
$$ {A}^{\dagger} C {B}^{\dagger} = \arg \min_{X} {\left\| C - A X B \right\|}_{F} $$
is the unique solution.
Note: This is an extension of the minimum $2$-norm least squares problem.
Hint: Use the SVDs of $A$ and $B$.

This is a homework problem, but I got stuck. After I do SVD for both $A$ and $B$, I discard the $0$ terms to get a slim SVD. Then I can factor out some things, but still don't see an explicit expression formulating. Could somebody guide me in the right direction?
Edit:
1."F norm": Frobenius norm
2. $A^+$ or $A^\dagger$: conjugate transpose, Moore–Penrose pseudoinverse

You don't need singular value decomposition here. Observe that $\pi_1:C\mapsto AA^+CB^+B$ and $\pi_2:X\mapsto A^+AXBB^+$ are orthogonal projections on their respective matrix spaces with respect to the usual inner products. Also,
$$
(C-\pi_1(C))\perp \pi_1(AXB)=AXB=A\pi_2(X)B.
$$
It follows that
$$
\|C-AXB\|_F^2 = \|\pi_1(C)-A\pi_2(X)B\|_F^2 + \|C-\pi_1(C)\|_F^2
$$
and the problem boils down to minimising $\|\pi_1(C)-A\pi_2(X)B\|_F$. As this quantity involves only the projected component $\pi_2(X)$ of $X$ but not its orthogonal component, in a least-norm solution, we must have $X=\pi_2(X)$.
Clearly, when $X=A^+CB^+$, we have $X=\pi_2(X)$ and $\|\pi_1(C)-A\pi_2(X)B\|_F$. Therefore $A^+CB^+$ is indeed a least-norm minimiser.
Now that the minimum possible value (i.e. zero) of $\|\pi_1(C)-A\pi_2(X)B\|_F$ is attained. If there is another least-norm minimiser $X$, we must have $X=\pi_2(X)$ and $\|\pi_1(C)-A\pi_2(X)B\|_F=0$. Thus we get
\begin{align}
A\pi_2(X)B &= \pi_1(C),\\
A^+A\pi_2(X)BB^+ &= A^+\pi_1(C)B^+,\\
\pi_2(\pi_2(X)) &= A^+AA^+CB^+BB^+,\\
X &= A^+CB^+
\end{align}
and we conclude that the least-norm solution is unique.
