A particular case that should help you think of a solution is when the sector is of the form $0\leq \text{arg}z \leq \pi$. In this case consider $g(z)=f(z)f(-z)$. Since $f$ is bounded, $g$ admits a continuous extension to the boundary of the circle that is $0$ on said boundary. The maximum modulus principle gives the result.
For any function $h\colon \Bbb C \to \Bbb C$ it is defined that
$$\frac{\partial h}{\partial z} = \frac12 \left( \frac{\partial h}{\partial x} - i \frac{\partial h}{\partial y} \right).$$
In particular, $u$ is such a function, so the above definition shows what $\frac{\partial u}{\partial z}$ means.
Now, a holomorphic function $f\colon U \subset \Bbb C \to \Bbb C$ has the property that, if $f = u + iv$, then
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$$
These are called the Cauchy-Riemann equations. Furthermore, it is not hard to see that also
$$f' = \frac{\partial f}{\partial x} \ \left( = \frac{\partial f}{\partial z} \right),$$
the second equation being an immediate consequence of the Cauchy-Riemann equations.
Using them, it is easy to see that if $f = u + iv$ is holomorphic, then $f'(z) = \frac{2 \partial u}{\partial z}$.
Now what you are to do is to take this function $g = \frac{2 \partial u}{\partial z}$ and show that it is holomorphic. To do this, the easiest way would be to show that $g$ satisfies the Cauchy-Riemann equations (they work in both ways!). This should not be hard, since this is basically just your assumption on $u$: We have $g = s + it$ where $s = \frac{\partial u}{\partial x}$ and $t = -\frac{\partial u}{\partial y}$, so, for example
$$\frac{\partial s}{\partial x} = \frac{\partial^2 u}{\partial x^2} = -\frac{\partial^2 u}{\partial y^2} = \frac{\partial t}{\partial y}.$$
The last step is to show that $g$ has an antiderivative $F$. This is true because $g$ is a holomorphic function on a simply connected region. To construct $F$, fix a point $z_0$ in the domain (i. e., the unit disc). For every $z$ in the domain define
$$F(z) = \int_{\gamma} g(w) dw, \quad \text{where $\gamma$ is a path from $z_0$ to $z$}.$$
That this actually works is not too hard to see if you know some theory behind it. Basically, we use the fact that the path integral of $g$ is equal for homotopic paths (in particular, it is $0$ on closed loops). Some basic calculations with the integral show that indeed $F' = g$.
Finally $F - f$ has zero derivative, so is constant.
Best Answer
OK. I tried sth different to rebuild my proof, with all the necessary details and explanation (I believe) contained. Here is the sketch and the complete version:
Sketch of the proof: first show that $f$ can be continuously extended to $0$ on the arc of the sector. Define a new function depending on $f$, possibly by rotating, such that it equals to $0$ on the boundary. Then apply the maximum modulus principle.
Proof: WLOG we assume that $\theta = 0$. $S:= \left\{ z\in\mathbb{D}: 0< \text{arg} z <\varphi\right\}$ is the sector. $f \rightrightarrows 0$ on $S$, i.e. $\forall \epsilon >0$, $\exists \delta >0$, s.t. $\forall z\in S$, and $\forall w\in \partial \mathbb{D}$, whenever $|z-w|<\delta$, we have $\left |f(z) \right |<\epsilon$.
Here we explains the uniformity. Basically we can imagine it dies to 0 very well on the boundary.Now we may extend $f$ to $\partial \mathbb{D}_{(0, \varphi)}$ (the arc of the sector) by letting $f = 0$.
For $\varphi/2$, there is a nature number $N\in \mathbb{N}$ s.t. $N \varphi /2 > 2\pi$.
We want to somehow cover the whole boundary of the disc so we may apply the max. mod. principle. However, the interval of argument is open. It is necessary and convenient to choose a number smaller than$\varphi$so that we can rotate it many times to cover all of the boundary with no holes at all.Consider $$F(z):= f(z)f(e^{-i\varphi/2}z)f(e^{-i2\varphi/2}z)\cdots f(e^{-iN\varphi/2}z)$$on $\mathbb{D}$. First, it is bounded on $\mathbb{D}$. Next, we show that it can be continuously extended to $\bar{\mathbb{D}}$: $\forall w\in \partial \mathbb{D}$, $\exists k\in \left\{ 0, 1, 2, \cdots, N\right\}$ s.t. $k\varphi/2 < \text{arg} w \leq (k+1)\varphi/2$. Now given $\epsilon >0$, $\exists \delta >0$ s.t. $|z-e^{-ik\varphi/2}w|<\delta$
(Rotate it back to where we start)implies $\left |f(z) \right |<\epsilon$; for the other numbers $j \neq k$ in $\left\{ 0, 1, 2, \cdots, N\right\}$, $\left |f(e^{-ij\varphi/2}z) \right |<B$ as $f$ is bounded. Hence $$\left |F(z) \right |= \prod_{j=0}^{N}\left |f(e^{-ij\varphi/2}z) \right |<B^N\epsilon.$$So we may define $F$ to be $0$ on $\partial \mathbb{D}$. Apply the maximum modulus principle. We conclude $F \equiv 0$ on $\bar{\mathbb{D}}$.
If $f \not \equiv 0$, then $f$ has at most countably many roots since it is analytic. We use this fact to conclude a contradiction. Now fix an arbitrary $\alpha$, $\forall re^{i\alpha}\in \mathbb{D}$, at least one of the term in $F(\alpha) = 0$ is zero. This gives a root for each $r \in (0, 1)$. But $(0, 1)$ is uncountable. Contradiction.
Therefore $f\equiv 0$. $\square $