Let $\mathcal{H}$ be a a vector space of finite dimension. Let $\mathcal{H_1}$ be a subspace of $\mathcal{H}$. Considering some vector $|\phi\rangle \in \mathcal{H}$ i need to show that there exists only one element of $\mathcal{H_1}$, say $|\phi_1\rangle$, that minimizes the norm $\|\phi - \phi_1\|$.
The norm is, of course, \begin{align} \|\phi - \phi_1\| &= \sqrt{\langle\phi-\phi_1|\phi-\phi_1\rangle} \\ &= \sqrt{ \langle\phi|\phi\rangle-\langle\phi|\phi_1\rangle-\langle\phi_1|\phi\rangle-\langle\phi_1|\phi_1\rangle }. \end{align}
I don't really know to progress any further to show that $\phi_1$ is unique or how to find $\phi_1$.
I seem to be stuck at this point so any help is much appreciated.
In a finite-dimensional inner product space $\mathcal H$ the orthogonal complement of a subspace is indeed a complement: $$ \mathcal H = \mathcal H_1 \oplus \mathcal H_1^\perp, \quad\text{where}\quad \mathcal H_1^\perp = \{\, |\psi\rangle \in\mathcal H \mid \langle x | \psi \rangle=0 \text{ for all $x\in\mathcal H_1$} \,\}. $$ Hence, given $|\phi\rangle\in\mathcal H$ you can always decompose it uniquely as $$ |\phi\rangle = |\phi_a\rangle+|\phi_b\rangle \quad\text{with $|\phi_a\rangle\in\mathcal H_1$ and $|\phi_b\rangle\in\mathcal H_1^\perp$}. $$ Now for $|\phi_1\rangle\in \mathcal H_1$ the calculation you started can be continued using $\langle\phi_1|\phi_b\rangle=0$ and $\langle \phi_a|\phi_b\rangle = 0$. \begin{align} \|\phi - \phi_1\|^2 &= \langle\phi-\phi_1|\phi-\phi_1\rangle \\ &= \|\phi\|^2 - \langle \phi|\phi_1\rangle - \langle\phi_1|\phi\rangle + \|\phi_1\|^2 \\ &= \underbrace{\|\phi_a\|^2 + \|\phi_b\|^2}_{\|\phi\|^2} - \langle\phi_a|\phi_1\rangle - \langle \phi_1|\phi_a\rangle + \|\phi_1\|^2 \\ &= \|\phi_b\|^2 + \|\phi_a-\phi_1\|^2. \end{align} Hence, the minimal value $\|\phi-\phi_1\|^2$ can obtain is $\|\phi_b\|^2$ and this happens exactly when $\phi_1=\phi_a$.