Why is $(A+BK, B)$ controllable when $(A,B)$ is, but $(C, A+BK)$ is not observable when $(C,A)$ is? I attempted to use the Popov-Belevitch-Hautus test to prove these, namely 
$$[A+BK - \lambda I, B]=[A-\lambda I, B]\begin{bmatrix}I & 0\\K & I \end{bmatrix}$$ 
which are both full rank and thus the pair is controllable. By the same idea, 
$$\begin{bmatrix}A+BK - \lambda I \\ C\end{bmatrix}=\begin{bmatrix}A-\lambda I & B\\ C & 0 \end{bmatrix}\begin{bmatrix}I\\K\end{bmatrix}$$ 
also should be full rank, so they should be observable... But from what I understand this is not always true. Can anyone show me proof of how it's true for controllability and not observability? This is in relation to open and closed loop systems, and the Popov-Belevitch-Hautus test is shown here.
I want to try and understand why its true for any K in the controllable case, but not the observable case
 A: An example:
$$A=\begin{bmatrix}0&1\\0&0\end{bmatrix},~~B=\begin{bmatrix}0\\1\end{bmatrix},~~C=\begin{bmatrix}1&1\end{bmatrix}$$
which is both controllable and observable. Now, we can select $K=\begin{bmatrix}-2&-3\end{bmatrix}$ to stabilize the closed loop system. But then $C(A+BK)=\begin{bmatrix}-2&-2\end{bmatrix}$, which shows that the closed loop system is not observable.
You might be interested how I came up with such an example, which is actually pretty straightforward. One of the eigenvectors of $A+BK$ is $\begin{bmatrix}-1&1\end{bmatrix}$, so I chose $C$ to be orthogonal to that vector for deliberately losing observability. This property is also obvious from your proof attempt.
It is not trivial to guess where the eigenvectors of $A+BK$ would end up, but the designer should guarantee that none of them are orthogonal to the rows of $C$ for keeping observability. Interestingly, the closed loop system can be modified to become observable. For example it would happen if $C=\begin{bmatrix}0&1\end{bmatrix}$. While this is mathematically true, in practice you need to measure all states to be able to use $K$, which can't be done if the system is not observable.
For the controllable case note that $BK$ is a matrix whose columns are linear combinations of columns of $B$, so you are not adding or subtracting anything from the controllable subspace. For a naive proof assume that $(A,B)$ is controllable but there exists a $K_1$ such that $(A+BK_1,B)$ is not controllable. But in this case I can select $\bar{K} = K - K_1$, which makes the closed loop matrix as $A+BK$. Since $K$ is a free parameter this means this system must also be controllable by the assumption.
