$V$ is finite-dimensional over $\Bbb{C}$ and the form $\langle \cdot , \cdot \rangle$ is Hermitian. $U$ is a subspace of $V$.
Show that $V = U \oplus U^\perp$
I've been able to show that $U \cap U^\perp = \{0\}$. I don't know how to approach the problem showing that every vector $v\in V$ can be written as $v = u + u'$, where $u, u'$ are in $U$ and $U^\perp$ respectively.
On
Choose any orthonormal basis $E=\{e_1 ,\ldots, e_k\}$ in $U$. Continue $E$ to a basis $E^+ = \{e_1 ,\ldots, e_k;\; v_{k+1},\ldots, v_n\} $ in $V$. Apply Gram-Schmidt to $E^+$. The first $k$ vectors in $E^+$ already are orthogonormal, and so Gram-Schmidt alters noting in them. The remaining $n-k$ vectors may change, and we get the orthonormal basis $E^{++} = \{e_1 ,\ldots, e_k;\; e_{k+1},\ldots, e_n\} $ in $V$. The vectors $e_{k+1},\ldots, e_n$ are orthogonal to $e_1 ,\ldots, e_k$, and so they all are in $U^\perp$. Present any $v\in V$ as the linear combination $v = (a_1 e_1 +\cdots+ a_k e_k)\;+\; (a_{k+1}e_{k+1}+\cdots + a_n e_n) = u+u'$. Clearly, $u \in U$ and $u' \in U^\perp$.
Let $\{e_1,\ldots,e_k\}$ be an orthonormal basis of the subspace $U$. For each $v\in V$, let$$P(v)=\sum_{j=1}^k\langle v,e_j\rangle e_j.$$Then$$(\forall v\in V):v=\overbrace{P(v)}^{\phantom{U}\in U}+\overbrace{\bigl(v-P(v)\bigr)}^{\phantom{U^\perp}\in U^\perp}.$$The fact that $v-P(v)\in U^\perp$ can be justified as follows: if $j\in\{1,2,\ldots,k\}$, then\begin{align}\bigl\langle v-P(v),e_j\bigr\rangle&=\left\langle v-\sum_{l=1}^k\langle v,e_l\rangle e_l,e_j\right\rangle\\&=\langle v,e_j\rangle-\langle v,e_j\rangle\\&=0.\end{align}Since $\{e_1,\ldots,e_k\}$ is a basis of $U$, this proves that $v-P(v)\in U^\perp$.