Let's say you have some density matrix, $\rho$, over a set of qubits $1,2,\ldots ,n$. (If you have a pure state $|\psi\rangle$, set $\rho=|\psi\rangle\langle\psi|$.) Let's say that the subsystem that we want the reduced density operator of is specified by the set of qubits $S$ (and all the others are $\bar S$). You have that
$$
\rho_S=\text{Tr}_{\bar S}(\rho),
$$
where $\text{Tr}_{\bar S}$ is the partial trace over the set $\bar S$. What does this mean in practice? Pick any orthonormal basis $\{|\phi_i\rangle\}_{i=1}^{|\bar S|}$ over $\bar S$, then you simply calculate
$$
\text{Tr}_{\bar S}(\rho)=\sum_{i=1}^{|\bar S|}\left(\langle \phi_i|_{\bar S}\otimes\mathbb{I}_S\right)\rho\left(| \phi_i\rangle_{\bar S}\otimes\mathbb{I}_S\right).
$$
Usually, you just pick the computational basis. So, in your example of
$$
\rho=\frac12\left(|00\rangle\langle 00|+|00\rangle\langle 11|+|11\rangle\langle 00|+|11\rangle\langle 11|\right),
$$
you pick the basis $\{|0\rangle,|1\rangle\}$, and you calculate
$$
\rho_A=(\mathbb{I}\otimes\langle 0|)\rho(\mathbb{I}\otimes|0\rangle)+(\mathbb{I}\otimes\langle 1|)\rho(\mathbb{I}\otimes|1\rangle)=\frac12(|0\rangle\langle 0|+|1\rangle\langle 1|)).
$$
To make this calculation a bit shorter, then it's worth noting that, over the subsystems $\bar S$ that you're taking the partial trace over, it's permutation invariant. So, you can pick up a ket from the left-hand side, and match it with a bra on the right-hand side (but make sure you always match terms from the same subsystem, and that that subsystem is in $\bar S$):
$$
\text{Tr}_B(\rho)=\frac12\left(|0\rangle\langle0|\langle0|0\rangle+|1\rangle\langle0|\langle0|1\rangle+|0\rangle\langle1|\langle1|0\rangle+|1\rangle\langle1|\langle1|1\rangle\right),
$$
and you can now simply read off this final result. It's always a good check, to make sure you haven't messed up too badly, to make sure that the $\rho_S$ that you output is (i) Hermitian, (ii) has trace 1.