Inconsistency between Bernoullis equation and continuity equation

Bernoulli equation is basically a energy conservation equation. But the version that you are using is incomplete. The term $$ \rho \frac{1}{2}V^2 $$ is a density of kinetic energy. The term $$ \rho g h $$ Is a gravitational potential density. But these cannot be the only terms.

EDIT

I decided to edit my answer to improve clarity. I disagree with the answers saying that we can't apply Bernoulli equation. In the Feynman lectures, we find a similar exercise, where Feynman apply Bernoulli equation for a container with a hole in the bottom. The difference there is that his container is open, here it is closed, but I don't see why we cannot apply energy conservation in this case.

To correct apply the principle of energy conservation, we first need to choose where in the fluid we will apply it and where will be the zero references for the potentials. The intuitive idea is to apply it in the point immediately before entering the pipe, in the bottom of first container. This will be my point $1$. This point will move to the end of the pipe, where we have the point $2$. Initially, let's not assume anything: we can have different velocities in both points $\vec V_1,V_2$, different pressures $p_2$, etc. I'll assume just same density and same displaced area of fluid $A$.

In a time $\Delta t$ after opening the pipe, the fluid in point $1$ will be displaced by $V_1\Delta t$ and in point $2$ $V_2\Delta t$. The continuity equation tell us that $$ \Delta M = \rho A V_1 \Delta t = \rho AV_2 \Delta t $$ So both points should have the same velocity. Since the density is the same and the area is the same, it means that the volume displaced is the same in both points, so $\Delta v_1 = \Delta v_2 =\Delta v$, where small $v$ here is for volume.

Now let's analyse the work done by pressure force. This is $$ W = p_1Av_1\Delta t - p_2Av_2\Delta t = (p_1-p_2) \Delta t. $$

By conservation of energy, the variation of energy in points 1 and 2 should be equal this work done. At point 1, using what I said in the beggining of this answer, the density of energy is $$ \frac{1}{2}\rho V_1^2 - \rho g h_1 $$ Where I placed the zero of the potential in the surface of height $h_1$, so in the bottom we have negative potential, and I wrote $V_1$ ignoring that we already discovered that the velocity in both points are the same, because I want to show that there is no inconsistency. We will have a similar expression for point 2. The increase of energy in $2$ is due to the increase in mass at this point. By same reasoning, the decrease of energy in $1$ is due to mass leaving this point. The total energy balance will be given by $$ W = \Delta M (E_2 - E_1) $$ Writting the full equation and dividing by $\Delta v$ we get $$ P_1 - P_2 = \rho \frac{V_2^2}{2} - \rho g h_2 - \rho \frac{V_1^2}{2} + mgh_1 $$ And this is our Bernoulli equation now. There is no contradiction with continuity equation, since we even used it to derive this equation. If I use the fact that both velocities are equal, we get $$ P_1 - P_2 = mgh_1 - mgh_2 $$ Which we know it is true. In the other hand, without saying both that both velocities are equal, but assuming $P = mgh$ in both points, we get $$ \rho \left ( \frac{V_1^2}{2}-\frac{V_2^2}{2}\right ) = 0 $$

Which implies that both velocities are equal. In resume, there is no contradiction between Bernoulli equation and continuity equation, since one is based on mass conservation and the other, energy conservation.

The continuity equation is based on the overall volume flow rate, assuming the average velocity across the pipe remains equal at both ends owing to the incompressibility of the fluid. The velocity change which Bernoulli's equation describes, represent localized change in velocity near the openings of the pipe. Not necessarily the same as the bulk velocity of the fluid contained in the pipe.

The missing link? Friction or viscous effects within the pipe. In real-life, there are losses owing to friction along the pipe, not captured in Bernoulli’s equation. Therefore, the energy available to convert into increased kinetic energy is reduced. Velocity difference predicted by Bernoulli’s equation would not show fully, making the average velocities along the pipe more equal as predicted by the continuity equation.

We can conclude that Bernoulli's equation focuses on energy conservation along a streamline, explaining a potential velocity difference between two points, while the continuity equation deals with bulk flow across cross-sections. So, both are valid in their own contexts, but here, real-world losses e.g. friction, reduce the effect that Bernoulli’s velocity difference predicts, leading to velocities being more uniform across the pipe as predicted by the continuity equation.

If we cut the problem into two parts: 2 containers with the communication tube (t) cut off, we have$$p_{atm}+\rho g h_{1}+\frac{1}{2}\rho V_{1}^{2}=p_{atm}+\frac{1}{2}\rho V_{t_{1}}^{2}$$ (from which we derive Torricelli's formula)

for the other container, we have $$\rho g h_{2}+\frac{1}{2}\rho V_{2}^{2}=\frac{1}{2}\rho V_{t_{2}}^{2}$$ i.e.if connected $$\rho g h_{1}+\frac{1}{2}\rho V_{1}^{2}-\frac{1}{2}\rho V_{t_{1}}^{2}=\rho g h_{2}+\frac{1}{2}\rho V_{2}^{2}-\frac{1}{2}\rho V_{t_{2}}^{2}$$ or $$\rho g h_{1}+\frac{1}{2}\rho V_{1}^{2}+\frac{1}{2}\rho V_{t_{2}}^{2}=\rho g h_{2}+\frac{1}{2}\rho V_{2}^{2}+\frac{1}{2}\rho V_{t_{1}}^{2}=cst$$ if $V_{1}=V_{2}$ , we have : $$\rho g (h_{1}- h_{2})=\frac{1}{2}\rho V_{t_{1}}^{2} -\frac{1}{2}\rho V_{t_{2}}^{2} \equiv\rho\int_{V_{t_{1}}>V_{t_{2}}} V_{t}dV$$

the two fluids will push each on its own side, what counts is the difference because the liquid at the same height in the two containers is in equilibrium..., we can write : $$ V_{t}=\sqrt{ 2g(h_{1}- h_{2})}\;\;\;\;\;\;\;\;\;\;(1)$$

The $dh_{1}$ variation of $h_{1}$ over time dt is therefore $$Adh_{1}=A_{t}V_{t}dt\;\;\;\;\;\;\;\;\;\;(2)$$ ( the continuity equation)

The conservation of fluid volume allows us to determine the final (common) level $h_{equilibrium}$: $$2Ah_{e}=A(h_{1}+h_{2})\;\;\;\;\;\;\;\;\;\;(3)$$

Relation (3) is used to express $h_{2}$ as a function of $h_{1}$ and $h_{e}$ in relation (1).

Integration of the differential equation (2) gives the law of variation of the edeal liquid height as a function of time.

Remeber that for an unsteady (i.e time dependent) irrotational flow Bernoulli says that $$ \frac {\partial \phi}{\partial t}+ \frac 12 |{\bf v}|^2 +h(\rho) $$ is constant along streamlines. Here $\phi$ is the velocity potential: ${\bf v}= \nabla \phi$, and $h$ is the specific enthalpy, which reduces to $P/\rho$ when the fluid is incompressible. The OP has omitted the $\partial \phi/\partial t$ in their discussion.