In general, an entangled state is one which cannot be decomposed as $\sum_{i}p_{i} \bigl(\rho_{i}^A\otimes\rho_{i}^B\bigr)$. But such an entangled state could still be mixed in principle.
How would one create a mixed entangled state? For instance, can one create such a state by tracing out one qubit or a subsystem of a pure state?
An example of a state which yeilds an entangled state when you trace one qubit out, is the 3-qubit "W" state: $$ \lvert W_3 \rangle = \tfrac{1}{\sqrt3} \Bigl( \lvert 100 \rangle + \lvert 010 \rangle + \lvert 001 \rangle \Bigr) $$ Taking the outer product with itself, we obtain $$ \lvert W_3 \rangle\!\langle W_3 \rvert = \tfrac{1}{3} \Bigl( \!\begin{aligned}[t]& \lvert 100 \rangle\!\langle 100 \rvert + \lvert 100 \rangle\!\langle 010 \rvert + \lvert 100 \rangle\!\langle 001 \rvert \phantom{\Big)} \\&+ \lvert 010 \rangle\!\langle 100 \rvert + \lvert 010 \rangle\!\langle 010 \rvert + \lvert 010 \rangle\!\langle 001 \rvert \\&+ \lvert 001 \rangle\!\langle 100 \rvert + \lvert 001 \rangle\!\langle 010 \rvert + \lvert 001 \rangle\!\langle 001 \rvert \Bigr) \end{aligned} $$ If we trace out the third qubit, we then obtain the state $$ \begin{align} \rho = \mathrm{tr}_3\Bigl(\lvert W_3 \rangle\!\langle W_3 \rvert\Bigr) &= \tfrac{1}{3} \Bigl( \lvert 10 \rangle\!\langle 10 \rvert + \lvert 10 \rangle\!\langle 01 \rvert + \lvert 01 \rangle\!\langle 10 \rvert + \lvert 01 \rangle\!\langle 01 \rvert + \lvert 00 \rangle\!\langle 00 \rvert \Bigr) \\&= \tfrac{1}{3} \lvert 00 \rangle\!\langle 00 \rvert + \tfrac{2}{3} \lvert \Psi^+ \rangle\!\langle \Psi^+ \rvert, \end{align} $$ where in particular $\lvert \Psi^+ \rangle = \tfrac{1}{\sqrt 2}\bigl( \lvert 01 \rangle + \lvert 10 \rangle \bigr)$ is a Bell state.
Note that $\rho$ is a rank-2 operator with different eigenvalues: any decomposition of $\rho$ as a convex combination of other density operators, can only involve terms whose eigenvectors are supported on $\mathrm{span}\,\{ \lvert 00 \rangle, \lvert \Psi^+ \rangle \}$. Any mixed tensor product density operator $\rho_A \otimes \rho_B$ has an eigenbasis consisting of two or more product states; but the only product state contained in $\mathrm{span}\,\{ \lvert 00 \rangle, \lvert \Psi^+ \rangle \}$ is the state $\lvert 00 \rangle$ itself. It follows that $\rho$ cannot be decomposed as a convex combination of products of possibly-mixed states, and is entangled.
More generally, if for $n > 1$ we define $\lvert W_n \rangle$ as the analogous $n$-term uniform superposition of standard basis states with a single 1, we may describe it as $$ \lvert W_n \rangle = \tfrac{\sqrt{n{-}1}}{\sqrt n} \lvert W_{n{-}1}\rangle\lvert0\rangle + \tfrac{1}{\sqrt n} \lvert00\cdots0\rangle\lvert1\rangle $$ so that $$ \mathrm{tr}_n\Bigl(\lvert W_n \rangle\!\langle W_n \rvert\Bigr) = \tfrac{n{-}1}{n} \lvert W_{n-1} \rangle\!\langle W_{n-1} \rvert + \tfrac{1}{n} \lvert 00\cdots0\rangle\!\langle 00\cdots 0\rvert $$ which for larger values of $n$ is closer and closer to being a pure entangled state, while still being mixed for any finite $n$.
A mixed separable state is written in the form $$ \rho=\sum_ip_i\sigma^A_i\otimes\sigma^B_i $$ where the $\sigma_i$ are valid density matrices on a single site.
The example you give, say $\rho=\frac12|00\rangle\langle00|+\frac12|11\rangle\langle 11|$ is exactly of this form. Specifically, $$ p_0=p_1=\frac12,\qquad \sigma^A_0=\sigma^B_0=|0\rangle\langle 0|,\qquad \sigma^A_1=\sigma^B_1=|1\rangle\langle 1|. $$
In general, it can be tricky to definitively prove that such a decomposition does not exist. However, in the case of two-qubits, there's an if and only if condition: the state is entangled if and only if its partial transpose is not non-negative (i.e. contains a negative eigenvalue).
The classic example is the Werner state, $$ \rho=\frac{1-p}{4}I+p|\psi\rangle\langle\psi| $$ where $|\psi\rangle$ is a two-qubit Bell state. This can be written out as a matrix $$ \rho=\left(\begin{array}{cccc} \frac{1-p}{4} & 0 & 0 & 0 \\ 0 & \frac{1+p}{4} & -\frac{p}{2} & 0 \\ 0 & -\frac{p}{2} & \frac{1+p}{4} & 0 \\ 0 & 0 & 0 & \frac{1-p}{4} \end{array}\right) $$ If we take the partial transpose of this, we get $$ \rho^T=\left(\begin{array}{cccc} \frac{1-p}{4} & 0 & 0 & -\frac{p}{2} \\ 0 & \frac{1+p}{4} & 0 & 0 \\ 0 & 0 & \frac{1+p}{4} & 0 \\ -\frac{p}{2} & 0 & 0 & \frac{1-p}{4} \end{array}\right) $$ This has a negative eigenvalue if $p/2>(1-p)/4$.
