The annihilator of an intersection is the sum of annihilators Given a subset $X$ of a vector space $V$, let $X^\circ$ be the annihilator of $X$, that is $X^\circ = \{y\in V^* | \; y(x)=0, \;\forall\; x\in X\}$, where $V^*$ is the dual space of $V$.

Question: If $M$ and $N$ are subspaces of an $n$-dimensional vector space $V$, then $(M\cap N)^\circ =M^\circ +N^\circ $.

It's easy to prove this result by using that $(M+N)^\circ =M^\circ \cap N^\circ $, which is easy to prove too. 

My doubt is: Is there a "direct way" to prove that $(M\cap N)^\circ =M^\circ +N^\circ $ (without make use of other equality)?

Thanks!
There is a reason why the first relation
$$(M+N)^\circ =M^\circ \cap N^\circ \tag1$$
is easier to prove than 
$$(M\cap N)^\circ =M^\circ +N^\circ\tag2$$
Indeed, (1) is true for arbitrary vector spaces. (The proof is short: a linear function annihilates $M+N$ if and only if it annihilates both $M$ and $N$.)

But (2) fails for general vector spaces (it holds for finite dimensional ones). More  specifically, the inclusion $(M\cap N)^\circ \supseteq M^\circ +N^\circ$ is true in general vector spaces, but not the reverse inclusion. So, any proof of (2) must use the finiteness of dimension somewhere. It cannot be a twin of the proof of (1).

(A gerw said, (2) can be made true in full generality by replacing $M^\circ +N^\circ$ with $\overline{M^\circ +N^\circ}$.)
You may work with orthonormal bases:
Let $\{u_i\} \cup \{v_i\} \cup \{w_i\} \cup \{x_i\}$ a basis in $X$, such that $\{u_i\}$ is a basis of $M\cap N$, $\{u_i\} \cup \{v_i\}$ of $M$ and  $\{u_i\} \cup \{w_i\}$ of $N$.

Then, $M^\circ$ is the linear hull of $\{w_i\} \cup \{x_i\}$, $N^\circ$ is the hull of $\{v_i\} \cup \{x_i\}$ and $(M \cap N)^\circ$ is the hull of $\{v_i\} \cup \{w_i\} \cup \{x_i\}$ (if I have done anything right).

Anyway, this is a mess. The easy proof you mentioned has also the advantage, that it works in infinite dimensions (in this case you have that the closure of $M^\circ + N^\circ$ is equal to $(M\cap N)^\circ$).