Let $(E,\langle \cdot,\cdot\rangle_1)$, $(F,\langle \cdot,\cdot\rangle_2)$ be two complex Hilbert spaces. We recall $$E \otimes F:=\left\{\xi=\sum_{i=1}^dv_i\otimes w_i:\;d\in \mathbb{N},\;\;v_i\in E,\;\;w_i\in F \right\}.$$
We endow $E \otimes F$, with the following inner product $$ \langle \xi,\eta\rangle=\sum_{i=1}^n\sum_{j=1}^m \langle x_i,z_j\rangle_1\langle y_i ,t_j\rangle_2, $$ for $\xi=\displaystyle\sum_{i=1}^nx_i\otimes y_i\in E \otimes F$ and $\eta=\displaystyle\sum_{j=1}^mz_j\otimes w_j\in E \otimes F$.
Let $(x_n)_n\subset E$ and $(y_n)_n\subset F$ such that $\displaystyle\lim_{n\to+\infty}x_n=x$ and $\displaystyle\lim_{n\to+\infty}y_n=y$. Why $$\displaystyle\lim_{n\to+\infty}x_n\otimes y_n=x\otimes y\;?$$
Thank you.
The map $T:E\times F\to E\otimes F$ mapping $(x,y)\to x\otimes y$ is bilinear and bounded. Hence, it is continuous. Note that $\lim_{n\to\infty} (x_n,y_n) = (x,y)\in E\times F$. Since $T$ is continuous, your statement follows.
We could also write $x_n=x+p_n$ and $y_n=y+q_n$ for two nullsequences $p_n$ and $q_n$, then \begin{align*} \|(x_n\otimes y_n)&-(x\otimes y)\|^2 = \langle (x_n\otimes y_n)-(x\otimes y), (x_n\otimes y_n)-(x\otimes y)\rangle \\&= \|x_n\|_1\|y_n\|_2 -2\langle x_n, x\rangle\langle y_n,y\rangle + \|x\|_1\|y\|_2 \\ &=\|x_n\|_1\|y_n\|_2 -2\langle x+p_n, x\rangle\langle y+q_n,y\rangle + \|x\|_1\|y\|_2 \\ &=\Bigl(\|x_n\|_1\|y_n\|_2 - \|x\|_1\|y\|_2\Bigr) - \langle p_n, x\rangle\langle q_n,y\rangle - \langle p_n, x\rangle\langle y,y\rangle - \langle x, x\rangle\langle y,q_n\rangle \end{align*} is a nullsequence because each summand is one and therefore $x_n\otimes y_n$ converges to $x\otimes y$.