$z = \sum_{i = 1}^n x_i \otimes y_i$ Let $U$ and $V$ be $K$-vector spaces, with $\{u_i\}_{i \in I}$ base of $U$ and $\{v_j\}_{j \in J}$ base of $V$. If $z \in U \otimes V$, prove there exists $y_i \in V$ unique such that $z = \sum_{i \in I} u_i \otimes y_i$.
The same can be proved for $x_j \in J$ unique such that $z = \sum_{j \in J} x_j \otimes v_j$.
Lastly, prove there exists $x_1, \dots, x_n \in U$ and $y_i, \dots, y_n \in V$, linearly independent such that $z = \sum_{i = 1}^n x_i \otimes y_i$.
I guess the first and second are easy. I did
$$z = (\sum_{i \in I} a_iu_i) \otimes v = \sum_{i \in I} u_i \otimes a_iv$$
Then $y_i = a_iv$.
The same for the second question. I don’t know if this is correct.
The last I don’t know how to prove.
 A: Your proof for the first (and second) question essentially establishes existence (see my comment below), but not uniqueness. You need to establish that if we have
$$
\sum_{i} u_i \otimes y_i = \sum_{i} u_i \otimes y_i',
$$
then we have $y_i = y_i'$ for all $i \in I$. Equivalently, we are given that $\sum_{i}u_i \otimes (y_i - y_i') = 0$, and we want to deduce that $y_i - y_i' = 0$ for all $i \in I$.
As in my answer on your previous, I would recommend that you show that if $y_i - y_i' \neq 0$ for some $i$, then $\sum_{i}u_i \otimes (y_i - y_i') \neq 0$. Do this by constructing a suitable linear map and appealing to the universal definition of the tensor product

I find the last question to be very tricky. Things are a bit easier if we are given that either $U$ or $V$ is finite dimensional, but even this case is a bit tricky.
The easiest method I can think of is to exploit the canonical identification of $U \otimes V$ with $\operatorname{Hom}(U^*,V)$ (where $U^*$ denotes the dual of $U$).  Because $z \in U \otimes V$, we can write $z = \sum_{i \in I} p_i \otimes q_i$ for some $p_i,q_i$ and finite indexing set $I$. Let $U_0 \subset U$ denote the (finite dimensional!) subspace of $U$ spanned by $\{p_i : i \in I\}$. Let $\Phi: \operatorname{Hom}(U_0^*,V) \to U_0 \otimes V$ denote the canonical isomorphism defined such that
$$
\Phi(vf(\cdot)) = f \otimes v.
$$
Let $\phi = \Phi^{-1}(z)$. Note that the (finite rank) map $\phi:U_0^* \to V$ can be decomposed into a "rank-factorization" $\phi = \phi_2 \circ \phi_1$ where $r$ is the rank of $\phi$, $\phi_1:U_0^* \to \Bbb F^r$, and $\phi_2:\Bbb F^r \to W$.  If $u_j \in U,v_j \in V$ are defined such that
$$
\phi_1(f) = (f(x_1),\dots,f(x_r)), \quad \phi_2(c_1,\dots,c_r) = c_1 y_1 + \cdots + c_r y_r,
$$
then it follows that
$$
\phi(f) = \sum_{j=1}^r f(x_j) y_j, \quad
\Phi(\phi) = \sum_{j=1}^r x_j \otimes y_j.
$$
