Angle between subspaces of inner product space

Solution 1:

(Let me work in real inner product spaces for simplicity, otherwise just take $\mathrm{Re}\langle x,y\rangle$ instead of $\langle x,y\rangle)$.

From the existence of $c>0$ such that $\|x+y\|^2\geq c(\|x\|^2+\|y\|^2)$ for all $x\in E$ and $y\in F$, we can write that:

$$\langle x,y\rangle\geq \frac{(c-1)}{2}(\|x\|^2 + \|y\|^2),\quad\forall x\in E,y\in F.$$

now note that in general $A^2 + B^2 \ge 2 AB$, so that you can write:

$$\langle x,y\rangle\geq \frac{(c-1)}{2}(\|x\| + \|y\|) \ge (c-1)\|x\|\|y\|,\quad\forall x\in E,y\in F.$$

Now notice we got $$\frac{\langle x,y\rangle}{\|x\|\|y\|} \ge (c-1)$$

By definition you are dealing with $\theta \in [0, \frac{\pi}{2}]$. To prove that $\theta > 0$ amounts to prove that $\theta \neq 0$, which is equivalent to $\cos(\theta) < 1$, that by definition means $$\sup\left\{\frac{|\langle x,y\rangle|}{\|x\|\|y\|}\colon x\in E,y\in F\right\} <1$$ that is for every $x \in E$ and $y \in F$ we have $$ |\langle x,y\rangle|<\|x\|\|y\| $$

Put them together and get for any $x,y$:

$$\|x\|\|y\| >|\langle x,y\rangle|\ge (c-1)\|x\|\|y\|$$

which is giving you $c < 2$. I would see the contradiction if the condition is "suppose there exists $c >2$", otherwise I do not see anything wrong with $\theta =0$.