Will the velocity be constant in a vertically oriented tube? 
The tube here is placed into a water stream moving with a velocity $v$
My query is will the velocity be constant throughout the pipe or only in some places?
I have conflicting opinions on this, the equation of continuity would say the velocity is constant through the pipe but Bernoulli's equation would say it slows down the higher it went.
Where have I flawed conceptually here?
The person who solved the question said the velocity at the blue points will be the same but in the portion above the level of water, it will not. This just confused me even more.
Link to the question although not necessary.
 A: 
My query is will the velocity be constant throughout the pipe

No. Flow will slowdown going up due to the stopping gravity force of water column. Check this scheme :

Water will raise-up to the height $h$ only due to dynamic pressure. Neglecting atmosphere pressure, maximum height until water can go-up can be calculated from hydro-static equilibrium which will be formed when water column of height $h$ gravitational pressure will be fully in balance with a flow dynamic pressure :
$$ P_d = P_g $$
Substituting dynamic pressure and gravitational pressure expressions gives :
$$ 1/2~ \rho u^2 = \rho gh $$
Solving for $h$, gives maximum possible water column height given flow velocity $u$ :
$$ h_{\text{max}} = \frac {u^2}{2g} $$
So, when flow velocity is zero,- $h=0$ too. If there exist maximum height to where water can raise (and hence flow velocity at that point $h_{\text{max}}$ also falling to zero too) - this indicates that water flow inside tube must decrease going-up.
A: The equation of continuity and Bernoulli's equation should always be satisfied (at least in ideal condition i.e. no viscosity and ...)
What you are forgetting is the pressure in Bernoulli's equation. The pressures at the beginning of blue line and at the end of the blue line are not the same. That explains how the velocity in the pipe can be the same everywhere. Let's write down the equation for the beginning of the blue line (point A) and the end of the blue line (point B)
$P_A + \frac{1}{2}mv^2 + \rho g h_A = P_B + \frac{1}{2}mv^2  + \rho g h_B$
$\frac{1}{2}mv^2$ cancels and we get,
$P_A + \rho g h_A = P_B  + \rho g h_B$
$P_A - P_B =  \rho g (h_B- h_A)$
because $h_B > h_A$ the LHS is positive. Thus, we can conclude that pressure at point A is larger that the pressure at the points B. This should make sense because the water moves from A to B so there should be some force from A toward B.
