Interpretation of directional depenedence in special relativity

Question

I am reading AP French's book, Special Relativity, because I have always wanted to understand the subject, and I have just attempted to answer the following question (1-22) in the book:

An electron moving with a speed $0.5c$ in the x direction enters a region of space in which there is a uniform electric field in the y direction. Show that the x component of the velocity of the particle must decrease.

My argument, which does not require making use of the specific number, $0.5c$, is as follows:

Let observer A see an electron moving in the positive x direction at speed $v_{x_{0}}$, and let observer B be moving with speed $v_{x_{0}}$ relative to observer A so that he is in the rest frame of the electron. According to observer B, when they enter the uniform electric field, which has magnitude E, the electron will begin moving in the y direction opposite the field under a constant force, $F=qE$.

After a time t, according to observer B, the particle must be moving in the y direction with velocity:

$$v_{y} = \frac{c}{(1+(\frac{m_{0}c}{Ft})^{2})^{\frac{1}{2}}} $$

Now I make a claim of whose veracity, I am not wholly convinced, but I cannot find a way out of it: because there is no relative motion in the y-direction between the observers, they must agree on the y-component of the electron's velocity.

So, to observer A, the electron is moving with squared velocity:

$$\vec{v} \circ \vec{v} = v_{x}^{2}(t) + v_{y}^{2}(t) = v_{x}^{2}(t) + \frac{c^{2}}{(1+(\frac{m_{0}c}{Ft})^{2})}$$

where I have allowed for a time-varying $v_{x}$. Taking limits as time goes to infinity, we can see that the second term tends to $c^{2}$, necessitating that the first term diminishes accordingly since $\vec{v}\circ\vec{v} < c^{2}$.

I have two questions:

First, is my reasoning correct?

Second, if the answer to my first question is yes, then suppose I have a third observer, C, and C begins at rest with respect to observer A, so she also sees the electron moving at 0.5c to the right. After some time T, when the electron has slowed appreciably in the x direction from the perspective of observer A, what x velocity would Observer C need to achieve relative to observer A to make the electron appear to be at rest in the x direction again? My suspicion is that the answer is 0.5c (otherwise, it would be inconsistent with observer B who started in the electron's rest frame and only ever sees it move in the y direction), and that the underlying physical mechanism that rectifies the seeming contradiction in x velocities is length contraction. I could also be dead wrong because for C to achieve any new velocity requires acceleration, which might break special relativity in a way that I haven't encountered yet.

UPDATE ONE: Thanks to @RogerJBarlow's comment, I began thinking about this problem in terms of energy and momentum. Because there is no external force acting in the x direction, momentum must be conserved in that direction. Since the electrostatic force acting in the y direction will accelerate the electron so that its speed increases, its inertial mass will increase, meaning that the x velocity must decrease if the x component of momentum is to be conserved.

Answer 1

Your claim of which you were not convinced about the veracity does not, alas, stand up (so your convictions were right.) A and B will agree about distances measured in the (transverse) $y$ direction, but their clocks run at different rates, so they disagree about $y$ velocities.

The way to approach the question (as in so many questions) is by considering energy and momentum rather than space and time. The electric field is transverse so it increases the energy of the electron but not the momentum component $p_x$. $p_x$ is given by $m_e \gamma v_x$: the increase in energy means $\gamma$ rises, so $v_x$ must fall.