If the covariance matrix is $\Sigma$, the covariance after projecting in $u$ is $u^T \Sigma u$. Why?

I read in this answer that:

If covariance matrix is $\Sigma$, the covariance after projecting in $u$ is $u^T \Sigma u$.

I fail to see this, how do I get the covariance of a set of points after projecting those points along the direction $u$ as a function of $u$ and $\Sigma$ ?

The covariance matrix for a vector quantity $x$ is $\langle xx^\top\rangle-\langle x\rangle\langle x^\top\rangle$. The covariance for the projection $u^\top x$ is

$$\langle u^\top xx^\top u\rangle-\langle u^\top x\rangle\langle x^\top u\rangle=u^\top\langle xx^\top\rangle u-u^\top\langle x\rangle\langle x^\top\rangle u=u^\top\left(\langle xx^\top\rangle-\langle x\rangle\langle x^\top\rangle\right)u\;.$$

The point is basically that you can pull $u$ out of all the expectation values because it's a constant.