Let $(H,\langle\cdot,\cdot\rangle))$ be a (real or complex) Hilbert space. Let $Y,Z\subseteq H$ be two closed linear subspaces. Let $\alpha(Y,Z)\in[0,\pi/2]$ be the unique number such that \begin{align} \cos \alpha(Y,Z) = \sup \{ |\langle y,z \rangle| : y\in Y, z\in Z, \|y\| = \|z\| = 1 \}. \end{align} Show that \begin{align} \alpha(Y,Z) > 0 \end{align} if and only if $\forall \varepsilon>0$, $\exists \delta>0$ such that \begin{align}\tag{1} \|y-z\|<\delta,\,y\in Y,\,z\in Z \quad \implies \quad \|y\|, \|z\| < \varepsilon. \end{align}
Is the following attempt OK?
My attempt:
$(\Rightarrow)$ Suppose that \begin{align} \alpha(Y,Z) > 0. \end{align} This is equivalent to that \begin{align} \gamma := \sup \{ |\langle y,z \rangle| : y\in Y, z\in Z, \|y\| = \|z\| = 1 \} < 1. \end{align} This implies that \begin{align} |\langle y,z \rangle| \leq \gamma\|y\| \|z\| \quad \forall(y,z)\in Y\times Z. \end{align} However, note that it $(y,z)\in Y\times Z$, then \begin{align} \|y-z\|^{2} &=\|y\|^{2} - 2\Re\langle y,z \rangle + \|z\|^{2} \\ &\geq\|y\|^{2} - 2|\langle y,z \rangle| + \|z\|^{2} \\ &\geq\|y\|^{2} - 2\gamma\|y\|\|z\| + \|z\|^{2} \\ &=(\|y\|-\|x\|)^2 +2(1-\gamma)\|y\|\|z\|. \end{align} But this implies (1).
$(\Leftarrow)$ We prove the contra positive. Suppose that \begin{align} \alpha(Y,Z) = 0. \end{align} This is equivalent to that \begin{align} \sup \{ |\langle y,z \rangle| : y\in Y, z\in Z, \|y\| = \|z\| = 1 \} = 1. \end{align} This implies that there exists $(y_{n},z_{n})\in Y\times Z$ such that $\|y_{n}\|=\|z_{n}\|=1$ for each $n\in\mathbb{N}$ and \begin{align} |\langle y_{n},z_{n} \rangle| \to 1. \end{align} W.l.o.g. we may assume that $|\langle y_{n},z_{n} \rangle| = \langle y_{n},z_{n} \rangle$ for each $n\in\mathbb{N}$. Note that \begin{align} \|y_n-z_n\|^2 &= \|y_n\|^2 - 2\Re(\langle y_{n},z_{n} \rangle) + \|z_n\|^2 \\ &=\|y_n\|^2 - 2\langle y_{n},z_{n} \rangle + \|z_n\|^2 \\ &=2(1-\langle y_{n},z_{n} \rangle) \to 0, \end{align} or simply \begin{align} \|y_n-z_n\| \to 0. \end{align} But this contradicts (1).
The $\impliedby$ part looks good. In the $\implies$ part you have shown that the product $\|y\|\|z\|$ is small, but this is not enough to conclude that both $\|y\|$ and $\|z\|$ are small.
To fix it, rearrange your argument like this:
\begin{align} \|y-z\|^{2} &\geq\|y\|^{2} - 2\gamma\|y\|\|z\| + \|z\|^{2} \\ &=(1 - \gamma)(\|y\|^2 + \|z\|^2) + \gamma(\|y\| - \|z\|)^2 \\ &\geq (1 - \gamma)(\|y\|^2 + \|z\|^2). \end{align}
In fact, proving $\implies$ only requires some Euclidean geometry.
Take any $\varepsilon > 0$; set $\delta = \varepsilon\sqrt{1-\gamma^2}$ will work, where $\gamma$ is the $\sup$ that you defined.
This is because both $\|y\|$ and $\|z\|$ do not exceed the circumdiameter $D$ of the triangle $0yz$.
By Euclidean geometry, we know that $D$ is equal to $\|y - z\|/\sin(\beta)$ where $\beta$ is the angle $\angle y0z$. Since we have $\cos(\beta) \leq \gamma$, it follows that $\sin(\beta) \geq \sqrt{1 - \gamma^2}$ and therefore $\|y\|, \|z\| \leq D < \delta / \sqrt{1 - \gamma^2} = \varepsilon$.