It is an interesting problem which is related to some recent work of mine. The reason for why I started to work on similar problems is because connections to a problem of Ramachandra on Dirichlet polynomials, connections to the nordic school of Hardy classes of Dirichlet series (Hedenmalm, Saksman, Seip, Olsen, Olofsson, Lindqvist and others), as well as universality questions for zeta-functions and their properties on the line Re(s)=1.
While my papers are not quite finished, I have put two early preprints on my homepage, On a problem of Ramachandra and approximation of functions by Dirichlet polynomials with bounded coefficients and On generalized Hardy classes of Dirichlet series. I have talked about some of these problems at analytic number theory conferences in India. Like in your paper I have considered Dirichlet series (it should be possible to obtain something like Theorem 2.1 in your paper by my method also, although I have not stated a direct analogue in my paper).
Now your problem in the question is rather easy for small $\omega$ so we will from now on assume that $\omega>1/2$. In fact if $\omega<1/2$, then $|f(0)|>1/2$ and $\int_0^K |\hat f(t)^2|dt \geq \min(1/10,K/10)$ (constants not chosen in an optimal way)
In my papers on Dirichlet series I have used a somewhat different method than you use in your paper, namely the Jensen inequality on the logarithmic integral in a half-plane. This method is applicable for the problem at hand. Lemma 7 in my paper ``On generalized Hardy classes of Dirichlet series'' can be used with $\sigma=0$ and $L(it)=\hat f(-t)$ and we obtain
$$\frac D \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {D^2+t^2} dt \leq \frac D \pi \int_{-\infty}^\infty \frac {\log^+ |\hat f (t)|} {D^2+t^2} dt - \log |\hat f(iD)|.
$$
For similar results see also Koosis - The logarithmic integral. (Remark Feb 16: The above inequality is an equality if the function is non-zero on a half plane. The inequality follows from Jensen's formula on a disc by mapping the half plane on the disc by the standard holomorphic bijection where $iD$ goes to $0$)
The reason why we can do this is that with the definition of the fourier-transform in your question it means that $ \hat f(z)$ will be a bounded analytic function in the half plane Im$(z) \geq 0$.
Now in this case we also have that $\log^+ |\hat f (t)|=0$ since $ |\hat f (t)| \leq 1$. Thus the inequality simplifies to
$$\frac D \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {D^2+t^2} dt \leq - \log |\hat f(iD)|.$$
It is not too difficult to see that for $\omega>1/2$
$$
|\hat f(i\omega)|= \left|\int_0^1 e^{i \phi(x)-\omega x} dx \right|>\frac {1} {10 \omega}.
$$
(The constant $10$ not chosen optimally). Thus we can choose $D=\omega$ and it is clear that
$$
\int_0^K \log^- |\hat f(t)| dt < \frac \pi {\omega} \left({\omega^2+K^2} \right) \frac {\omega} \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {\omega^2+t^2} dt
$$
From these estimates we see that
$$
\frac 1 K \int_0^K \log^- |\hat f(t)| dt< \frac {\pi(\omega^2+K^2)}{\omega K} \log (10 \omega).
$$
Now we can use the Jensen inequality
$$
\exp\left(\frac 1 K \int_0^K \log |\hat f(t)| dt\right)< \sqrt{\frac 1 K \int_0^K |\hat f(t)|^2 dt}
$$
We get the lower bound
$$
K \left(\frac 1 {10 \omega} \right)^{2\pi (\omega^2+K^2)/(K \omega)} \leq \int_0^K |\hat f(t)|^2 dt
$$
for $\omega>1/2$. If $c>2 \pi$ and $\omega/K$ is sufficiently large this gives a lower bound
$$\omega^{-c \omega/K} \leq \int_0^K |\hat f(t)|^2 dt$$
which is weaker than your expected $e^{-c \omega/K}$. At least we have an explicit lower bound.
Updated Feb 16: In the case where both $\omega$ and $K$ are large but still $\omega>K$ this can be improved by the following trick. Let $g$ be the convolution of $\hat f$
with a non negative test-function $\Phi(t/K)$, such that $\hat \Phi(0)>0$ where $\Phi$ has support on $[0,1/2]$ . Then use Jensen's inequalities on the function $g$ instead of $\hat f$ as above. The advantage with this is that it then follows that $|\hat g(iw)| \gg K/\omega$ and thus we can get the lower bound (by using Jensen's inequality w.r.t the L^1-norm instead of the L^2-norm.)
$$(\omega/K)^{-c \omega/K} \leq \frac 1 K \int_0^{K/2} |g(t)| dt$$
for some constant $c>0$. Since
$$ g(t)=\int_0^t \Phi((t-x)/K) \hat f(x) dx$$ it is clear by the triangle inequality that
$$\frac 1 K \int_0^{K/2} |g(t)| dt = \frac 1 K \int_0^{K/2} \left|\int_0^t \Phi((t-x)/K)\hat f(x) \right| dx \leq $$
$$\leq \frac 1 K \int_0^{K/2} |f(x)| dx \int_0^{K/2} |\Phi(x/K)| dx \leq c \int_0^{K/2} |\hat f(x)| dx $$
The inequality
$$K^{-1} (\omega/K)^{-c \omega/K} \leq \int_0^{K/2} |\hat f(t)|^2 dt$$
follows by the Cauchy-Schwarz inequality for some constant $c>0$.
This formula just use involves dimensionless quantity $\omega/K$ as expected. Since the function $E(K)$ is increasing in $K$ it gives the lower bound $E(K) > C_0 K^{-1}>0$ for $1 \leq \omega \leq K$ for some absolute constant $C_0$.
For $\mathbb{R}$. Suppose f is our compactly supported function and g(x) is its Fourier transform. Since f is compactly supported, $\hat{f} = g$ is the restriction to $\mathbb{R}$ of an entire function g(z) by the Paley-Wiener theorems. Since g is entire and vanishes on an open set, $g \equiv 0$. The proof of this last fact (weakening the assumption to vanishing on a set with an accumulation point) uses that $\mathbb{C}$ is connected which is of course directly related to $\mathbb{R}$ being connected.
I expect that you knew this proof, but maybe you accidentally overlooked where connectedness was used. Or more likely, this proof didn't explain what you had in mind and you want a more general proof for $\mathbb{R}^n$. I can't currently do that. Instead, I have another idea which focuses on a different aspect than connectedness, but seems to be related.
In connection with the analogous statement for polynomials. A polynomial can only have finitely many zeroes over a field is proved via a complexity argument using that infinity > finite. Analytic functions, i.e. the completion of polynomials over $\mathbb{C}$ can have infinitely many zeroes, but uncountably many zeroes implies the analytic function is identically 0. So it seems that a set that has a limit point is more complex (in terms of complexity) than a countable set. I'm thinking the complexity argument should be interpreted in terms of density in topology - no finite subset of a $\mathbb{N}$ is dense in the discrete topology or any open subset of the co-finite topology on $\mathbb{N}$. Similarly for $\mathbb{R}$ and $\mathbb{C}$.
I hope this is helpful. This is an interesting question and I'll think more about it.
Best Answer
Yes, there is one such example: $u \equiv 0$.
The answer above is not facetious! That $u$ is in fact the only example (modulo measure zero modifications).
By Titchmarsh's theorem, if $u\in L^2(\mathbb{R})$ and its Fourier support is on the positive real line, $u$ must be equal to the trace of some holomorphic function $F$ defined on the upper half plane.
If $u$ itself further vanishes on the left half line, which has positive measure, by the Luzin-Privalov Theorem the function $F$ must vanish identically. Hence the only function satisfying your condition is identically zero.