The continuity equation only means that the mass flow rate in equals the mass flow rate out. It does not mean that that flow rate never changes; flow in and flow out can both change simultaneously.
For a given pressure drop (pressure upstream minus pressure downstream), the flow rate is proportional to the 4th power of radius, according to the Hagen–Poiseuille equation. If you decrease radius, flow in AND flow out decrease. The continuity equation is still obeyed.
So if there is a decreased radius of a blood vessel, either flow rate decreases, or pressure increases to maintain the orginal flow rate, or there is a combination of pressure increase and flow decrease.
In the form you've stated it ($A_1 v_1 = A_2 v_2$), the continuity equation is only holds for incompressible fluids. So what you've found is that the type of accelerated flow you're describing cannot happen for an incompressible fluid in a pipe of uniform cross-section. The more general form for the continuity equation is based on conservation of mass (i.e., mass per time entering = mass per time exiting), and states
$$
Q_m = \rho_1 A_1 v_1 = \rho_2 A_2 v_2,
$$
where $Q_m$ is the mass flow rate (i.e., mass per time). This means that if the fluid increases in velocity, it must decrease in density.
An analogy here would be cars on a highway. Suppose you have a highway that leads from the center of a small town out into the country. Suppose further that the drivers are all perfectly safe drivers and obey the two-second rule, i.e., the cars pass a given point in the highway at a rate of one car every two seconds. If the speed limit in the town is low, then the cars will be more closely spaced, since two seconds corresponds to less distance. Thus, the density of cars is higher at this point. When the cars get out of town and the speed limit increases, the cars get further apart in distance (since two seconds now corresponds to a longer distance), and so the density decreases.
Bernoulli's equation, meanwhile, doesn't hold so simple a form for compressible fluids. Rather, you have to define a pressure potential $w(P)$ from the equation of state $\rho(P)$:
$$
\frac{1}{2} v^2 + gy + w(P) = \mathrm{const.},
$$
where
$$
w(P) = \int \frac{\mathrm dP}{\rho(P)}.
$$
For the case of an "incompressible" fluid, $\rho(P) = \rho$ is a constant, $w(P) = P/\rho$, and the familiar form of Bernoulli's Law is recovered. But for a compressible fluid, the equation may look quite different.
Best Answer
You are looking at the volume only in time, but you are forgetting how volume is calculated in the first place:
$$dV = A \cdot ds$$
where $A$ is the cross-section area and $ds$ is the infinitesimally small tube length for which you are calculating the infinitesimally small volume:
$$ds = v \cdot dt \tag 1$$
where $v$ is the fluid velocity and $dt$ is the infinitesimally small time. The Eq. (1) is key to understanding where your reasoning is wrong - for the same time $dt$ the fluid covers longer distance through smaller cross-section area, which we prove via mass flow.
The infinitesimally small mass flowing into the tube across $A$ in time $dt$ is:
$$dm = \rho \cdot dV = \rho \cdot A \cdot v \cdot dt$$
where $\rho$ is the fluid density which is constant for incompressible fluids.
Since the mass is not stored in the system, for any two points in the tube we have
$$dm_1 = dm_2 \quad \text{or} \quad A_1 v_1 = A_2 v_2$$
Continuity is not violated since water speed at the two points is different. In other words, for infinitesimal small time $dt$ the fluid covers greater distance $ds$ for smaller cross-section area $A$. You have simply neglected the effect $dt$ has on $ds$ and assumed that $ds$ is constant.