If $A$ is compact and $B$ is Lindelöf space , will be $A \cup B$ Lindelöf

The product of two $KC$ spaces need not be $KC$. Let $\Bbb Q^*$ be the one-point compactification of the rationals; then $\Bbb Q^*$ is $KC$, but $X=\Bbb Q^*\times\Bbb Q^*$ is not. It's well-known that $\Bbb Q^*$ is $KC$. To see that $X$ is not, let $\Delta=\{\langle x,x\rangle:x\in\Bbb Q^*\}$. Then $\Delta$ is homeomorphic to $\Bbb Q^*$, so $\Delta$ is compact; I'll show that $\Delta$ is not closed in $X$ and hence that $X$ is not $KC$.

Let $p=\langle\infty,0\rangle\in X$. For each $\epsilon>0$ let $I_\epsilon=(-\epsilon,\epsilon)\cap\Bbb Q$. For each compact subset $K$ of $\Bbb Q$ and $\epsilon>0$ let $B(K,\epsilon)=(\Bbb Q^*\setminus K)\times I_\epsilon$, and let $\mathscr{B}$ be the family of all such $B(K,\epsilon)$; $\mathscr{B}$ is a local base at $p$. Fix $B(K,\epsilon)\in\mathscr{B}$. Clearly $I_\epsilon\setminus K\ne\varnothing$, so let $y\in I_\epsilon\setminus K$; then $\langle y,y\rangle\in\Delta\cap B(K,\epsilon)$, and since $B(K,\epsilon)$ was arbitrary, it follows that $p\in(\operatorname{cl}_X\Delta)\setminus\Delta$ and hence that $\Delta$ is not closed.