Why is a density matrix an orthogonal projector?

Suppose I have a density matrix like $$\rho = \frac{1}{2}[I + \hat{n}\vec{\sigma}]$$.

The claim is that $$\rho$$ is an orthogonal projector for the state $$|+\rangle$$ in an arbitrary direction $$\hat{n}$$.

How do I convince myself of this claim?

If a matrix is a projector, it squares to itself. So, you just have to verify that $$\rho^2=\rho$$. You'll need the assumption that the length of the Bloch vector is 1.

Suppose I have a density matrix like $$\rho = \frac{1}{2}[I + \hat{n}\vec{\sigma}]$$.

Explicitly, we have $$\rho = \frac{1}{2}\left(\begin{matrix} 1+n_z & n_x - in_y\\ n_x + in_y & 1 - n_z \end{matrix}\right)$$

The claim is that $$\rho$$ is an orthogonal projector for the state $$|+\rangle$$ in an arbitrary direction $$\hat{n}$$.

How do I convince myself of this claim?

You can convince yourself based on the well known Bloch sphere expression for a ket in the direction $$\hat n$$: $$|\psi\rangle = \cos(\theta/2)|0\rangle + e^{i\phi}\sin(\theta/2)|1\rangle\;,$$ where $$\hat n = (\sin(\theta)\cos(\phi), \sin(\theta)\sin(\phi), \cos(\theta))\;.$$

Then the density operator is: $$\hat \rho = |\psi\rangle\langle\psi|$$ and, for example the (0,0) matrix element of the above operator is: $$\rho_{00} = \langle 0|\psi\rangle\langle\psi|0\rangle = \cos(\theta/2)^2 = \frac{1+\cos(\theta)}{2}\;,$$ which is the (0,0) element of your original matrix, since: \begin{align} \rho &= \frac{1}{2}\left(\begin{matrix} 1+n_z & n_x - in_y\\ n_x + in_y & 1 - n_z \end{matrix}\right)\\ &= \frac{1}{2}\left(\begin{matrix} 1+\cos(\theta) & \sin(\theta)(\cos(\phi)) -i\sin(\phi))\\ \sin(\theta)(\cos(\phi)) +i\sin(\phi)) & 1 - \cos(\theta) \end{matrix}\right)\;. \end{align}

And so on.