Do intersections commute with direct sum? This is just a basic linear algebra question without that much context to it.  I'm wondering if the following identity holds for vector spaces:
$$ (A \oplus  B)  \cap C  = (A \oplus 0 )\cap C  + (0 \oplus B)\cap C. $$

My intuition tells me it's always true, but I could be wrong.
This is so wrong...

Let $A=Span(1,0), B=Span(0,1), C=Span(1,1)$.
Then
$$(A\oplus B)\cap C=\mathbb{R}^2\cap C=C$$ and
$$(A\cap C)\oplus (B\cap C)=0\oplus 0 = 0.$$
No, vector spaces don't verify that:
$$(A \oplus  B)  \cap C  = (A\cap C)  \oplus (B\cap C).$$
To prove it, we can find a counterexpample:
Consider the vector subspaces $\langle\begin{pmatrix} 1 \\ 0 \end{pmatrix}\rangle$, $\langle\begin{pmatrix} 0 \\ 1 \end{pmatrix}\rangle$ and $\langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle$. Notice that
$$\left(\langle\begin{pmatrix} 1 \\ 0 \end{pmatrix}\rangle\oplus \langle\begin{pmatrix} 0 \\ 1 \end{pmatrix}\rangle\right)\cap \langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle=\mathbb{R}^2\cap \langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle=\langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle,$$
but
$$\left(\langle\begin{pmatrix} 1 \\ 0 \end{pmatrix}\rangle  \cap \langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle\right)\oplus\left(\langle\begin{pmatrix} 0 \\ 1 \end{pmatrix}\rangle  \cap \langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle\right)=0\oplus 0=0,$$
and clearly $\langle\begin{pmatrix} 1 \\ 1 \end{pmatrix}\rangle\neq 0$.