To check that a set of $n\times n$ matrices $\subset \mathbb{R}^{n^2}$ is compact, it is probably easiest to check that it is closed and bounded.
I'm going to give an example that is not any of the three you have in the problem, but will hopefully help you with them. Let $S$ be the collection of $n\times n$ symmetric matrices. We can then ask: is $S$ closed? Is $S$ bounded? Is $S$ compact?
I claim that $S$ is in fact closed. To see this, suppose that $M_n$ is a sequence of elements in $S$, i.e., a sequence of symmetric matrices, and suppose $M_n$ converges to a matrix $M$. To prove that $S$ is closed, we must show that $M$ is also a symmetric matrix. Write $M_n = (a_{ij}^{(n)})$, and $M = (a_{ij})$. The fact that $M_n\to M$ says exactly that $a_{ij}^{(n)}\to a_{ij}$ as $n\to \infty$ for each $i$ and $j$. But $M_n$ is symmetric, so $a_{ij}^{(n)} = a_{ji}^{(n)}$, so we get $a_{ij} = \lim_n a_{ij}^{(n)} = \lim_n a_{ji}^{(n)} = a_{ji}.$ This proves $M$ is symmetric, and that $S$ is closed.
I claim next that $S$ is not bounded. $S$ is bounded if and only if the entries of all the matrices in $S$ are bounded uniformly, so to prove my claim I need only find elements of $S$ with arbitrarily large entries. For each $c\in \mathbb{R}$, the matrix $M$ all of whose entries are $c$ lies in $S$, proving that the matrices in $S$ do not have uniformly bounded entries. This proves $S$ is not bounded.
Now is $S$ compact? $S$ is compact if and only if it is closed and bounded. Since it is not bounded, it is not compact. I hope this helps.