To clarify your confusion:
Weierstrass M-test states one and only one thing:
Given a sequence of functions $f_k(x)$ defined on $E\subseteq\mathbb{R}$, the series $\sum f_k(x)$ converges uniformly if there exists a sequnece of reals $M_k$ such that $|f_k|\le M_k$ for each $k$ and $\sum M_k$ converges.
Note here that $M_k$ must not depend on $x$. This only means that if you found such sequence $M_k$ then the series uniformly converges. It says nothing about the series if you have found some $M_k$ whose series does not converge.
In particular, this test cannot be used to prove that some series does not converge uniformly.
Now consider
$$
f_k(x)=\frac{1}{kx+2}-\frac{1}{(k+1)x+2}
$$
Consider the partial sum
\begin{align*}
S_n(x)=\sum_{k=0}^{n} f_k(x)&=\left(\frac{1}{2}-\frac{1}{x+2}\right)+\cdots+\left(\frac{1}{nx+2}-\frac{1}{(n+1)x+2}\right)\\
&=\frac{1}{2}-\frac{1}{(n+1)x+2}
\end{align*}
the series $\sum f_k(x)$ converges uniformly if and only if $S_n$ converges uniformly.
Now $S_n(0)=0$ for all $n$. So $S_n(0) \to 0$, and $S_n(x) \to \frac{1}{2}$ for $0<x\le 1$. So define
$$
S(x)=
\begin{cases}
0 & \text{if} \quad x=0 \\
\frac{1}{2} & \text{if} \quad 0<x\le 1
\end{cases}
$$
Then $S_n(x)\to S(x)$. Also, $S_n(0)-S(0)=0$ for any $n$. Now
$$
m_n:=\sup_{x\in [0,1]} |S_n-S|=\sup_{x\in (0,1]} \left\lvert -\frac{1}{(n+1)x+2} \right\rvert=\frac{1}{2} \not\to 0
$$
So $S_n$ does not converge uniformly.
We know
$$I=\int_0^\infty \frac{\sin x}{x}\, dx >0.$$
Suppose there exists $M>0$ such that $b>M$ implies
$$|\int_b^\infty \frac{\sin (rx)}{x}\, dx |<\frac{I}{2}$$
for all $r\ge 0.$ Letting $x=y/r$ then gives
$$|\int_{rb}^\infty \frac{\sin y}{y})\, dy |<\frac{I}{2}$$
for all $r\ge 0.$ Now let $r\to 0^+$ to arrive at $I\le I/2,$ contradiction.
Best Answer
The series (2) is not uniformly convergent for $x \in (0,\infty)$.
We have
$$\begin{align}\sup_{x \in (0,\infty)}\left|\sum_{k=n+1}^\infty \frac{x}{x^2 + k^2}|\sin k|\right| &\geqslant \sup_{x \in (0,\infty)}\sum_{k=n+1}^{2n} \frac{x}{x^2 + k^2}|\sin k| \\ &\geqslant \sup_{x \in (0,\infty)}\frac{x}{x^2 + 4n^2}\sum_{k=n+1}^{2n} |\sin k|\\ &\geqslant\frac{n}{n^2 + 4n^2}\sum_{k=n+1}^{2n} |\sin k|\end{align}$$
Using $|\sin k| \geqslant \sin^2k = (1- \cos 2k)/2$, it follows that
$$\sup_{x \in (0,\infty)}\left|\sum_{k=n+1}^\infty \frac{x}{x^2 + k^2}|\sin k|\right| \geqslant \frac{1}{5}\left(\frac{1}{2}- \frac{1}{2n}\sum_{k=n+1}^{2n} \cos 2k\right)$$
The RHS converges to $1/10 \neq 0$ as $n \to \infty$ since the sum over $\cos 2k$ is bounded for all $n$. Hence, the convergence is not uniform for $x \in (0,\infty)$ by violation of the uniform Cauchy criterion.