I have worked out the following auxiliary statement. Can anyone confirm that the proof is correct and consistent? And would there be a way to show or present this in a simpler or more elegant way? Or did I make it too complicated? Would be very grateful for tips and criticism!
EDIT: Unfortunately, I failed to prove that $\kappa$ does not depend on the choice of $v_1$ and $v_2$, but on their norm and inner product. Are there any hints/suggestions on how this could be easily seen from my proof? It seems not obvious to me.
Let $(H, \langle \cdot, \cdot \rangle_H)$ be a Hilbert space with induced norm $\| \cdot \|_H$. Let $v_1, v_2 \in H$ satisfy \begin{equation} 1 + \langle v_1, v_2 \rangle_H \neq 0. \tag{1}\label{technicalcond} \end{equation} Then there exists a real constant $\kappa = \kappa(\|v_1\|_H, \|v_2\|_H, \langle v_1, v_2 \rangle_H) > 0$ such that \begin{equation} \|u + v_1 \langle u, v_2 \rangle_H\|^2_H \geq \kappa \|u\|^2_H \tag{2}\label{technicalresult} \end{equation} for all $u \in H$.
Proof.
We assume $v_1 \neq 0$ and $v_2 \neq 0,$ as the other case is trivial. We consider the following real-valued functional
\begin{equation}
I(u) := \|u + v_1 \langle u, v_2 \rangle_H\|^2_H = \Vert u \Vert_H^2 + 2 \langle u,v_1\rangle_H \langle u,v_2\rangle_H + \Vert v_1 \Vert^2_H \langle u,v_2\rangle_H^2 \tag{3}
\end{equation} which is nothing but the left-hand side of \eqref{technicalresult}. In the first step we will show that the quadratic functional $I$ is positive definite. Obviously $I(u)$ is non-negative and $I(0)=0.$ So we have to demonstrate that $I(u)=0$ implies $u=0.$ By the positive definiteness of norms, if
\begin{equation*} I(u) = \|u + v_1 \langle u, v_2 \rangle_H\|_H^2 = 0,\end{equation*} then \begin{equation} u + v_1 \langle u, v_2 \rangle_H = 0, \tag{4} \label{keineIdee} \end{equation}
which is equivalent to
\begin{equation} u = - v_1 \langle u, v_2 \rangle_H. \tag{5}\label{keinPlan} \end{equation}
Inserting \eqref{keinPlan} into \eqref{keineIdee} yields
\begin{equation*} - v_1 \langle u, v_2 \rangle_H - v_1 \langle v_1 \langle u, v_2 \rangle_H, v_2 \rangle_H = 0 \end{equation*}
and consequently, \begin{equation*} v_1 \langle u, v_2 \rangle_H + v_1 \langle u, v_2 \rangle_H \langle v_1 , v_2 \rangle_H = v_1 \langle u, v_2 \rangle_H (1 + \langle v_1 , v_2 \rangle_H) = 0. \end{equation*}
By \eqref{technicalcond} we conclude that $\langle u, v_2 \rangle_H = 0$ and \eqref{keinPlan} implies $u=0$.Since the linear space $\mathrm{span}\{v_1, v_2\} $ is a finite-dimensional subspace of $H,$ it is closed in $H,$ which allows us to decompose the Hilbert space $H$ as
$ H = \mathrm{span}\{v_1, v_2\} \oplus \mathrm{span}\{v_1, v_2\}^{\perp}.$ This means that every $u \in H$ can be uniquely represented as
$ u = u_{12} + u^{\perp}$ with $u_{12} \in \mathrm{span}\{v_1, v_2\}$ and $u^{\perp} \in \mathrm{span}\{v_1, v_2\}^{\perp}.$ This insight gives us
\begin{align*}
I(u) &= I(u_{12} + u^{\perp}) \\
&= \Vert u_{12} + u^{\perp} \Vert^2 + 2 \langle u_{12} + u^{\perp}, v_1 \rangle \langle u_{12} + u^{\perp}, v_2 \rangle + \Vert v_1 \Vert^2 \langle u_{12} + u^{\perp}, v_2 \rangle \\
&= \Vert u_{12} \Vert^2 + \Vert u^{\perp} \Vert^2 + 2 \langle u_{12}, v_1 \rangle \langle u_{12}, v_2 \rangle + \Vert v_1 \Vert^2 \langle u_{12}, v_2 \rangle \\
&= I(u_{12}) + \Vert u^{\perp} \Vert^2.
\end{align*}
If we restrict $I$ on the compact set $ \mathrm{span}\{v_1, v_2\} \cap \{ u \in H \, \colon \, \Vert u \Vert_H = 1 \} $
$= \{ u \in \mathrm{span}\{v_1, v_2\} \, \colon \, \Vert u \Vert_H = 1 \},$ it follows that the continuous functional $I$ attains its minimum $\tilde{\kappa} \geq 0,$ which cannot be equal to zero by definiteness.
For the case $u \in \mathrm{span}\{v_1, v_2\} \setminus \{0\}$ let us consider $\tilde{u} := \frac{u}{\Vert u \Vert_H} \in \{ u \in \mathrm{span}\{v_1, v_2\} \, \colon \, \Vert u \Vert_H = 1 \}.$ The previous statement and the homogeneity of $I$ imply
\begin{equation*}
\frac{1}{\Vert u \Vert_H^2} I(u)= I\left(\frac{u}{\Vert u \Vert_H} \right ) = I(\tilde{u}) \geq \tilde{\kappa}.
\end{equation*}
This altogether yields for the general case $u \in H,$
\begin{equation*}
I(u) = I(u_{12}) + \Vert u^{\perp} \Vert^2 \geq \tilde{\kappa} \Vert u \Vert_H^2 + \Vert u^{\perp} \Vert^2 \geq \underbrace{\min \{1, \tilde{\kappa} \}}_{=:\kappa} (\Vert u \Vert_H^2 + \Vert u^{\perp} \Vert^2) = \kappa \Vert u \Vert_H^2.
\end{equation*} $ \Box $
As I said in my earlier comment, your proof was almost complete, you just needed to show that $\kappa$ was independent on the choice of $v_1$ and $v_2$, and I just saw what to do to tie this neatly into your proof.
We can see in your proof that $\kappa$ already only depends on the values of $I$ on the intersection of the span of $v_1$ and $v_2$ with the unit sphere. What can be done is then showing that, when $(\|v_1\|_H, \|v_2\|_H, \langle v_1, v_2\rangle_H) = (\|v_1\|_H, \|v_2\|_H, \langle v_1, v_2\rangle_H)$, we have (indexing the $I$s temporarily, and denoting by $\mathbb{K}$ the scalar field): $$\forall (\alpha, \beta) \in \mathbb{K}^2,\quad I_{v_1, v_2}(\alpha v_1 + \beta v_2) = I_{v'_1, v'_2}(\alpha v'_1 + \beta v'_2) \tag{1}$$ and that: $$\forall (\alpha, \beta) \in \mathbb{K}^2,\quad \|\alpha v_1 + \beta v_2\|_H = 1 \Leftrightarrow \|\alpha v'_1 + \beta v'_2\|_H = 1 \tag{2}$$ I'll show $(2)$ first since it's fairly straightforward. We get, looking at the norms: $$\|\alpha v_1 + \beta v_2\|_H^2 = \langle \alpha v_1 + \beta v_2, \alpha v_1 + \beta v_2\rangle_H = \alpha^2 \|v_1\|_H^2 + \beta^2 \|v_2\|_H^2 + 2\alpha \beta\mathrm{Re}\langle v_1, v_2\rangle_H$$ Therefore we even have the condition $\|\alpha v_1 + \beta v_2\|_H = \|\alpha v'_1 + \beta v'_2\|_H$, which of course implies $(2)$.
I will leave the details for $(1)$ as an exercise, but it's essentially the same idea, starting from here: $$\begin{split}&I_{v_1, v_2}(\alpha v_1 + \beta v_2)\\ &= \|\alpha v_1 + \beta v_2\|^2_H + 2\langle \alpha v_1 + \beta v_2, v_1\rangle_H \langle \alpha v_1 + \beta v_2, v_2\rangle_H + \|v_1\|_H^2 \langle \alpha v_1 + \beta v_2, v_2\rangle_H^2\end{split}$$ and developing the inner products to make the desired quantities appear and only them (along with $\alpha$ and $\beta$ of course).
Combining $(1)$ and $(2)$ then provides the independence of $\kappa$ on the choice of $v_1$ and $v_2$ since the functionals $I$ will be equal on the respective sets $\kappa$ was chosen from.