An inequality for the dimension of the sum of subspaces

The answer to question 2. is No for every $n\geq4$.

Here is a counterexample for $n=4$:

Let $\mathbf{k}$ be a field, and let $M=\mathbf{k}^{6}$.

Let $U_{1}=\left\{ \left( a,b,c,d,e,f\right) \in M\ \mid\ a=b\text{ and }d=e\right\} $.

Let $U_{2}=\left\{ \left( a,b,c,d,e,f\right) \in M\ \mid\ b=c\text{ and }e=f\right\} $.

Let $U_{3}=\left\{ \left( a,b,c,d,e,f\right) \in M\ \mid\ c=d\text{ and }f=a\right\} $.

Let $U_{4}=\left\{ \left( a,b,c,d,e,f\right) \in M\ \mid \ a+c+e=b+d+f\right\} $.

Then, your inequality becomes

$6\leq4+4+4+5-2-2-2-3-3-3+1+1+1+1-1$,

which is false.

To get counterexamples for any $n\geq4$, extend this by setting $U_{5} =U_{6}=\cdots=U_{n}=0$.

Yes, for $n=3$.

I assume known the equality in the case $n=2$, so $\dim (U+V)= \dim(U) +\dim(V) - \dim (U\cap V)$.

Using this twice, we have $\dim(U +V + W) = \dim(U +V) + \dim (W) - \dim((U +V) \cap W) = \dim(U ) + \dim(V) + \dim (W) - \dim(U \cap V) - \dim((U +V) \cap W) $.

As $(U+V) \cap W \supset (U \cap W) + (V \cap W) $ we have

$\dim((U +V) \cap W) \ge \dim ( (U \cap W) + (V \cap W))$

Then $ \dim ( (U \cap W) + (V \cap W)) = \dim(U \cap W) + \dim(V \cap W) - \dim((U \cap W) \cap (V \cap W))$.

As $(U \cap W) \cap (V \cap W)= U \cap V \cap W$, the claim follows.