compactness and fundamental group of $SL(n,\mathbb{C})$ could any one help me to prove the heading?
$SL(n,\mathbb{C})$ is closed I can prove only, what are the other tools I need?
$SL(n,\mathbb{C})$ connected?simply connected?

The case $n=1$ is trivial, so assume $n\ge 2$. It is easy to see that $SL(n,\mathbb{C})$ is not compact, just look at diagonal matrices with entries $\lambda, \lambda^{-1}, 1, \ldots,1$, and let $\lambda \to \infty$. It is equally easy to see that it is connected, just deform the Jordan normal form to the identity matrix. For the fundamental group you need a little more sophisticated tools, see the other answer.
