Yes, it is compact: Consider a finite-dimensional complex Hilbert space and the space of all (bounded) linear operators on $H$, denoted by $\mathcal B(H)$. The set of all density matrices is $$\mathcal S(H):=\left\{\rho\in \mathcal B(H)\,|\, \mathrm{Tr}\rho=1, \, \rho \geq 0\right\} \tag 1\quad .$$
As a subset of a finite-dimensional Hilbert space, we have to show that $\mathcal S(H)$ is closed and bounded. We can define an inner product on $\mathcal B(H)$ by $\langle \cdot,\cdot\rangle:\mathcal B(H)\times\mathcal B(H)\to \mathbb C$ via $\langle A,B\rangle:=\mathrm{Tr}A^\dagger B$ for $A,B\in \mathcal B(H)$.
Now note that $\mathcal S(H)$ is closed: Consider a converging sequence of density operators $(\rho_n)_n$ with limit $\rho \in \mathcal B(H)$. We have to show that $\rho$ is a density operator. To this end, recall that inner products are continuous and so we have that $$0\leq \mathrm{Tr}\rho_nA \to \mathrm{Tr}\rho A\quad ,\tag 2$$
for all positive semi-definite $A\in \mathcal B(H)$, which implies that $\rho \geq 0$. Choosing $A=\mathbb I_H$, the identity operator on $H$, also shows that $\mathrm{Tr}\rho=1$, so $\rho\in\mathcal S(H)$.
Boundedness of $\mathcal S(H)$ in turn follows from the fact that for all $\rho \in S(H)$, it holds that
$$\mathrm{Tr}\rho^2 \leq 1 \tag 3\quad .$$