Do we always assume principal values converge? Consider this simple textbook real integral 
$$I = p.v.   \int_{-\infty}^{\infty} \frac{\cos x}{x^2 + 4} dx  = \lim_{\rho \to \infty} \int_{-\rho}^{\rho} \frac{\cos x}{x^2 + 4} dx$$
Now in many books it can be split as such 
$$I = p.v. \int_{-\infty}^{\infty} \frac{e^{ix}}{2(x^2 + 4)} dx +  p.v. \int_{-\infty}^{\infty} \frac{e^{-ix}}{2(x^2 + 4)} dx \quad (*)$$
where we are integrating over the real line so its imaginary component $y = 0$ and so $x$ is completely real. Then Jordan's lemma can be applied to show the circular part tends to $0$.
But in order to write $(*)$, don't we have to assume $p.v. \int_{-\infty}^{\infty} \frac{e^{ix}}{2(x^2 + 4)} dx$ has to exist first? Otherwise how do we justify splitting the original integral
$$p.v. \int_{-\infty}^{\infty} \frac{e^{ix} + e^{-ix}}{2(x^2 + 4)} dx$$
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Consider $$\int_{0}^{\infty} \frac{\cos x - 1}{x^2} dx = -\pi/2.$$ This can be found via $Re \frac{1}{2}\int_{-\infty}^{\infty} \frac{e^{ix}-1 }{x^2} dx$. But this is only true $\iff$
$$\lim_{\rho \to \infty}\int_{-\rho}^{\rho} \frac{e^{ix}-1 }{x^2} dx = \lim_{\rho \to \infty}\int_{-\rho}^{\rho} \frac{\cos x - 1}{x^2} dx + i\lim_{\rho \to \infty}\int_{-\rho}^{\rho} \frac{\sin x}{x^2} dx$$
But in all the answers I am reading from the manual, they never bothered to show one part ($i\lim_{\rho \to \infty}\int_{-\rho}^{\rho} \frac{\sin x}{x^2} dx$) converges. They just assume either the imaginary/real part converges and take the real/imaginary part. 
And I threw $i\lim_{\rho \to \infty}\int_{-\rho}^{\rho} \frac{\sin x}{x^2} dx$ into Mathematica and it gives no value. 
Now I know the original integral converges because it is a textbook problem, but there is something fundamentally wrong in immediately assuming we can split the integral like that.
 A: 
Do we always assume principal values converge?

No.

But in order to write $(*)$, don't we have to assume $p.v. \int_{-\infty}^{\infty} \frac{e^{ix}}{2(x^2 + 4)} dx$ has to exist first?

Yes.  However, that's what you should show in the next steps.  In essence, we show the justification for being able to split the sum at the same time we evaluate the principal values of the summands.  If you want it to follow more logically, you could reverse the order in your solution.
Then, paraphrased from the comments, we have:

Why are we considering the principal value integral?

Typically, the solution method as a contour integral only proves that the principal value integral exists, not the more general improper integral.  If we are able to show that the improper integral itself exists, then evaluating the principal value integral allows us to conclude the full improper integral has the same value.

In reference to your latest example:
The short answer is that $\frac{\sin x}{x^2}$ is an odd function with one singularity at $x=0$, so $$\text{p.v.}\!\int_{-\infty}^\infty \frac{\sin x}{x^2}\,dx = 0$$ does exist.
In the theory, we always have that $z_n \in \mathbb{C}$ converges to $a+bi$ ($a,b \in \mathbb{R}$) if and only if $\Re(z_n), \Im(z_n) \in \mathbb{R}$ each converge to $a$ and $b$ respectively.
In your latest example, the solution would have shown that $$\text{p.v.}\!\int_{-\infty}^\infty \frac{e^{ix}-1}{x^2}\,dx$$ exists and equals $-\pi$ (I haven't checked whether this is the right value, but it sounds plausible).  As a result of the above theorem we may conclude that both the following principal value integrals converge with the following values:
$$\text{p.v.}\!\int_{-\infty}^\infty \frac{\cos x - 1}{x^2}\,dx = -\pi \\ \text{p.v.}\!\int_{-\infty}^\infty \frac{\sin x}{x^2}\,dx = 0$$
To rephrase my answer to the second highlighted question above for the present context, the method used in the solution to evaluate the one complex-valued integral proves both the imaginary and real parts converge with the appropriate values.
As for why your Mathematica input gave no output for the limit
$$\lim_{\rho\to\infty} \int_{-\rho}^\rho \frac{\sin x}{x^2}\,dx$$
it's because this integral doesn't exist.  This, however, isn't a problem since this isn't how the principal value integral is evaluated.  Specifically, in this case,
$$\text{p.v.}\!\int_{-\infty}^\infty \frac{\sin x}{x^2}\,dx = \lim_{\rho\to\infty \\ \epsilon\to 0^+} \left(\int_{-\rho}^{-\epsilon}\frac{\sin x}{x^2}\,dx + \int_\epsilon^\rho \frac{\sin x}{x^2}\,dx\right)$$
