Why $\displaystyle\lim_{n\to+\infty}x_n\otimes y_n=x\otimes y\;?$

You can prove it directly using Squeeze theorem for sequences.

$$\begin{split} 0 \leqslant &\|x_n\otimes y_n - x\otimes y\| \hspace{110pt}(\text{subtract and add }x_n\otimes y)\\ = & \|x_n\otimes y_n - x_n \otimes y + x_n\otimes y - x\otimes y\| \hspace{28pt}(\text{triangle inequality})\\ \leqslant&\|x_n\otimes y_n - x_n \otimes y\| + \|x_n\otimes y - x\otimes y\| \hspace{18pt}(\text{grouping})\\ = &\|x_n\otimes (y_n - y)\| + \|(x_n - x)\otimes y\| \hspace{44pt}(\text{cross norm } \|x\otimes y\|=\|x\|_1\|y\|_2)\\ = &\|x_n\|_1\|y_n-y\|_2+\|x_n-x\|_1\|y\|_2\rightarrow 0 \end{split}$$

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 $x\otimes y$.