$T$ be a finite rank operator of rank $n$. There exist orthonormal set $\{u_1,\ldots,u_n\}$ such that $\{Tu_1,\ldots,Tu_n\}$ is linearly independent and $$Th=\sum_{i=1}^n\langle h,u_i\rangle Tu_i$$
Let's start with an orthonormal basis $\{v_1',\ldots,v_n'\}$ of $\text{Ran}(T)$. Then $v_i'=Tu_i'$ for some $u_i'\in H$. Then $\{u_1',\ldots,u_n'\}$ is linearly independent. By Gram-Smith orthogonalization there is a orthonormal set $\{u_1,\ldots,u_n\}$ such that $\text{Span}\{u_1,\ldots,u_n\}=\text{Span}\{u_1',\ldots,u_n'\}$ and therefore $\{Tu_1,\ldots,Tu_n\}$ is linearly interdependent.
Now extend the orthonormal set $\{u_1,\ldots,u_n\}$ to an orthonormal basis $\{u_i\}\cup\{e_\alpha\}$ of $H$. Then for any $h\in H$, $h=\sum\limits_{i=1}^n\langle h,u_i\rangle u_i+\sum\limits_{\alpha}\langle h,e_\alpha\rangle e_\alpha$. This implies $Th=\sum\limits_{i=1}^n\langle h,u_i\rangle Tu_i+\sum\limits_{\alpha}\langle h,e_\alpha\rangle Te_\alpha$.
So we just need to have $Te_\alpha=0\ \forall\alpha$ or $\text{Ran}(T^*)=\text{Ker}(T)^\perp=\text{Span}\{u_1,\ldots,u_n\}$.
Can anyone help me with an idea to finish the proof? Thanks for your valuable assistance in advance.
I suggest to use some of the same ideas you have hanging around, but in a slightly different way. As Ryszard mentioned, you also need that $T$ is bounded.
You can write $H=(\ker T)^\perp\oplus \ker T$ (the boudnedness of $T$ guarantees that $\ker T$ is closed and that $T^*$ exists and is defined everywhere). On $(\ker T)^\perp$, $T$ is injective: indeed, if $h\in(\ker T)^\perp$ and $Th=0$, we know that $h\in(\ker T)^\perp=\operatorname{ran} T^*$, so $h=T^*k$ for some $k\in H$. Then $$ 0=\langle Th,k\rangle=\langle h,T^*k\rangle=\|h\|^2, $$ so $h=0$. Then $\dim(\ker T)^\perp=\dim\operatorname{ran}T$. Now simply choose $u_1,\ldots,u_n$ an orthonormal basis of $(\ker T)^\perp$. Given any $h\in H$, we can write $$ h=\sum_{j=1}^n\langle h,u_j\rangle\,u_j\,+\,k $$ for some $k\in\ker T$. Then $$ Th=\sum_{j=1}^n\langle h,u_j\rangle\,Tu_j. $$