If $M=U\cup V$, $U$ and $V$ have finite dimensional cohomology then $U\cap V$ has finite dimensional cohomology..

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I need some help with the following: Let $M$ be a differentiable manifold such that $$\textrm{dim}(H^k(M))<\infty$$ for every $k=0, \ldots, n$ where $H^k(M)$ is the $k$-th De Rham cohomology group of $M$. Furthermore, suppose $M=U\cup V$ where $U$ and $V$ are open sets such that $$\textrm{dim}(H^k(U)), \textrm{dim}(H^k(V))<\infty$$ for every $k=0, \ldots, n$. I want to use Mayer-Vietoris sequence to show $\textrm{dim}(H^k(U\cap V))<\infty$ for every $k=0, \ldots, n$.

Partial Solution: Consider the exact sequence $$\ldots \longrightarrow H^k(U)\oplus H^k(V)\stackrel{j^*}{\longrightarrow}H^k(U\cap V)\stackrel{\partial}{\longrightarrow} H^{k+1}(M)\stackrel{j^*}{\longrightarrow}H^{k+1}(U)\oplus H^{k+1}(V)\rightarrow \ldots$$

Since $\partial:H^k(U\cap V)\rightarrow H^{k+1}(M)$ is linear is true that $$H^k(U\cap V)\simeq\textrm{ker}(\partial)\oplus \textrm{img}(\partial).$$ But the above sequence is exact hence $\textrm{ker}(\partial)=\textrm{im}(j^*)$ and $\textrm{img}(\partial)=\textrm{ker}(i^*)$ so that $$H^k(U\cap V)\simeq \textrm{im}(j^*)\oplus \textrm{ker}(i^*).$$ Since $j^*$ is linear and $H^{k}(U)\oplus H^k(V)$ is finite dimensional $\textrm{im}(j^*)$ is also finite dimensional.

Problem: For finishing the proof it suffices showing $\textrm{ker}(i^*)$ is finite dimensional but I don't know how to do it, can anyone help me? Thanks..

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Be careful, by the way, with splitting everywhere. But for dimensionality issues, you're ok. You went one step too far. $\text{img}\,\partial$ is finite-dimensional, as it's a subspace of a finite-dimensional space.