Is $T$ (defined below) a distribution? 
How to show that 
  $$ 
\langle T, \varphi\rangle = \int\limits_{0}^{+\infty} \frac{\varphi(x)-\varphi(0)}{x^{3/2}} dx,\quad \varphi \in C_0^\infty(\mathbb{R}),
$$ 
  is a distribution?

I would first rewrite the definition of $T$, using  $ \varphi(x)-\varphi(0) = \int_0^x \varphi'(t) dt $. But how to get around the fact that $ \int_{0}^{+\infty} x^{-3/2} dx $ is divergent, how to "tame" this integral in order to show that $ \lvert \langle T, \varphi\rangle \rvert \leq C \: \text{sup}_{K} \lvert \varphi'(x) \rvert $ for all $ \varphi \in C_0^\infty(K)$?

As $\varphi \in C_0^\infty(\mathbb{R})$, you can write
$$\varphi(x) - \varphi(0) = x \varphi^\prime(0) + x \epsilon(x)$$ with
$\lim\limits_{x \to 0} \epsilon(x) = 0$. As $\int_0^1 \frac{dx}{x^{1/2}}$ converges, $\int\limits_0^1 \frac{\varphi(x)-\varphi(0)}{x^{3/2}} dx$ converges.
Then for $x$ large enough, say $ x >M >0$ , $\varphi(x)$ vanishes. And as $\int_M^{\infty} \frac{\varphi(0)}{x^{3/2}} \ dx$ converges, $\langle T, \varphi \rangle$ is well defined.
