Prove that every $G$-invariant subspace of $V \otimes U$ has the form $V \otimes U_{0}$. Knowing the following theorem:

Theorem 4. Let $T$ be an irreducible complex representation of the group $G$ in the space $V$, and let $I$ be the trivial representation of $G$ in the space $U$. Then every minimal subspace $W \subset V \otimes U$ invariant under the representation $TI$ has the form $V \otimes u_0$, where $u_0 \in U$.


How can I prove that:

Every $G$-invariant subspace of $V \otimes U$ has the form  $V \otimes U_{0}$, where $U_{0}$ is a  subspace of $U$.

Could anyone help me please?
 A: The following works whenever $V$ has dimension $< |\mathbb{C}|=\mathfrak{c}$, so it is not a complete answer.
I will use the Jacobson density theorem, as stated for instance on the wikipedia page with the same name (I will use left modules instead of right modules but that doesn't matter of course):

The Jacobson Density Theorem. Let $U$ be a simple left $R$-module, $D = \operatorname{End}\left(_R U\right)$, and $X \subset U$ a finite and $D$-linearly independent set. If $A$ is a $D$-linear transformation on $U$ then there exists $r \in R$ such that $A\left(x\right) = rx$ for all $x \in X$. (Isaacs, Theorem 13.14, p. 185.)

Indeed $V$ is an irreducible representation of $G$, hence a simple $\mathbb{C}G$-module.  
Now by Schur's lemma, with our hypothesis on the dimension of $V$, $End_{\mathbb{C}G}(V) = \mathbb{C}Id$ (the usual form of Schur's lemma assumes finite dimensions, but actually you can make it work if the dimension is strictly smaller than the size of the field - if you don't know about that I can add a few words for that).
Therefore, using the notations of wikipedia, that is $D=End_{\mathbb{C}G}(V) = \mathbb{C}Id$, a subset $X$ of $V$ is $D$-linearly independent if and only if it is $\mathbb{C}$-linearly independent, and a $D$-linear map is simply a linear map: we will use this.
Let $(e_i)_{i\in I}$ be a basis for $V$.
Let's now get to the proof : let $W\subset V\otimes U$ be a $G$-invariant subspace, and let $\displaystyle\sum_{i\in J}e_i\otimes u_i\in W$, with $J$ a finite subset of $I$, $u_i\in U\setminus\{0\}$ for $i\in J$. 
Then for all $\alpha\in \mathbb{C}G$ as $W$ is $G$-invariant and $U$ has a trivial $G$-action, we get that $\displaystyle\sum_{i\in J}(\alpha e_i)\otimes u_i \in W$. 
Let $i\in J$ and $j \in I$, and let $f_{ij}:V\to V$ be defined by sending every $e_k$ to $0$ except for $e_i$, which it sends to $e_j$. Then $f_{ij}$ is linear, hence $D$ linear. Hence, by the Jacobson density theorem, since $X=\{e_j, j\in J\}$ is linearly independent, hence $D$-linearly independent, we get that there is $\alpha \in \mathbb{C}G$ such that $f_{ij}(x)= \alpha x$ for $x\in X$. In particular $e_j\otimes u_i =\displaystyle\sum_{k\in J}\alpha e_k\otimes u_k\in W$.
Therefore $V\otimes u_i\subset W$ for all $i$. So now if you take $U_0 = \{u\in U\mid \exists i\in I, e_i\otimes u \in W\}$. 
By the Jacobson density theorem again, it follows that $U_0$ is a subspace of $U$, moreover by the above proof it's clear that $V\otimes U_0 = W$. 
I don't know if the result is true when $V$ has arbitrary dimension, and if it is I don't know how to prove it (yet). I do suspect that the result you were expected to prove had some kind of finiteness assumptions. If you have some finiteness assumptions ($G$ finite, or more generally $V$ finite dimensional), the proof becomes much easier
