This answer tries to get things straight, but yes, there's a tiny piece missing in step $(4)$.
$\qquad(1)$: $\forall x \in\mathbb R$, the series $\sum_{n\geq 1}(-1)^n\frac{x^2}{n^2+x^2}$ converges absolutely.
Proof: We have that
$$\sum_{n\geq 1}\frac{x^2}{n^2+x^2}=x^2\,\sum_{n\geq 1}\frac{1}{n^2+x^2}\leq x^2\sum_{n\geq 1}\frac{1}{n^2}=\frac{x^2\pi^2}6.\qquad\qquad\square$$
$\qquad(2)$: $\forall x \in \mathbb R$, the series $\sum_{n=1}^{\infty}\left(\frac{x}{\left(2n\right)^{2}+x^{2}}-\frac{x}{\left(2n-1\right)^{2}+x^{2}}\right)$ converges and we have $x\sum_{n=1}^{\infty}\left(\frac{x}{\left(2n\right)^{2}+x^{2}}-\frac{x}{\left(2n-1\right)^{2}+x^{2}}\right)=\sum_{n\geq 1}(-1)^n\frac{x^2}{n^2+x^2}$.
Proof: Consider the partial sums
$$S_m=\sum_{n=1}^m(-1)^n\frac{x^2}{n^2+x^2}$$
and
$$T_m=x\sum_{n=1}^{m}\left(\frac{x}{\left(2n\right)^{2}+x^{2}}-\frac{x}{\left(2n-1\right)^{2}+x^{2}}\right).$$
By $(1)$, $S_m$ converges as $m\to\infty$.
It suffices to note that $T_m=S_{2m}$. $\square$
$\qquad(3)$: $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)=0$
Proof: Expand
$$\pm\left (\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right )-\frac1{n^2}$$
and verify that the result is negative for all real $x$ and positive integers $n$.
Conclude that the following estimate holds:
$$\left |\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right |\leq\frac{1}{n^2}\quad\forall n\in\mathbb{Z^+},\forall x\in\mathbb{R}$$
It then follows from the Weierstrass M-test that $\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)$ converges uniformly and absolutely on $\mathbb{R}$.
Since uniform convergence holds, we have
\begin{align}
&\lim_{x\to\infty}\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)\\
=&\sum_{n=1}^{\infty}\lim_{x\to\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)\\
=&\sum_{n=1}^{\infty}(2-1-1)=\sum_{n=1}^{\infty}0=0
\end{align}
which concludes the proof. $\square$
$\qquad(4)$: $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)$ and $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right)$ both exist, and they are equal.
Partial Proof: It follows from $(3)$ and the algebra of limits that
$$\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)
=
\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right)$$
provided both limits exist.
$\qquad(5)$: $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}x^{2}}{n^{2}+x^{2}}=-1/2$
Proof: For each $x\in\mathbb{R}$ we have
$$\sum_{n=1}^{m}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)\\
=
-\frac{x^2}{1+x^2}
-\sum_{n=1}^{m}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right)
-\frac{x^2}{\left(2m+1\right)^{2}+x^2}.$$
Letting $m\to\infty$, we conclude that for all $x\in\mathbb{R}$
$$\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)
=
-\frac{x^2}{1+x^2}
-\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right),$$
where the series on the LHS converges by $(2)$, and similarly the RHS series also converges.
Now, let $L=\lim_{x\to\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)$.
Letting $x\to\infty$ in the equality above and applying $(4)$, we get
$$L=-1-L\iff L=-1/2.$$
The claim follows from noting that $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}x^{2}}{n^{2}+x^{2}}=L$ as per $(2)$. $\square$
EDIT: We can use the integral test to arrive at the answer straight after step $(2)$.
Indeed, for each $x\in\mathbb{R}$ let $a_x(n)=\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}$, so we are interested in $\lim_{x\to\infty}\sum_{n\geq 1}a_x(n)$.
Observe that $a_x(n)<0$ whenever $x\neq 0$ and $n\geq1$, so we may apply the integral test to $\sum_{n\geq 1}-a_x(n)$.
We will have that
$$\int_1^\infty-a_x(t)\,dt\leq\sum_{n=1}^{\infty}-a_x(n)\leq -a_x(1) + \int_1^\infty-a_x(t)\,dt.$$
On the one hand, $-a_x(1)=\frac{x^2}{1+x^{2}}-\frac{x^2}{4+x^{2}}$.
On the other,
\begin{align}
\int_1^\infty-a_x(t)\,dt
&=\int_1^\infty\frac{x^2}{\left(2t-1\right)^{2}+x^{2}}-\frac{x^2}{\left(2t\right)^{2}+x^{2}}\,dt
\\&=-\frac{x}{2}\cdot\left[\arctan\left(\frac{1-2t}x\right)+\arctan\left(\frac{2t}x\right)\right]_{t=1}^\infty
\end{align}
The brackets are simply
$\left[\lim_{t\to\infty}\left(\arctan\left(\frac{1-2t}x\right)+\arctan\left(\frac{2t}x\right)\right)-\arctan\left(\frac{-1}x\right)-\arctan\left(\frac2x\right)\right]$,
and since we have $\lim_{t\to\infty}\arctan\left(\frac{1-2t}x\right)=-\pi/2$ and $\lim_{t\to\infty}\arctan\left(\frac{2t}x\right)=\pi/2$, it follows that
$$\int_1^\infty-a_x(t)\,dt=\frac{x}{2}\left(\arctan\left(\frac{-1}x\right)+\arctan\left(\frac2x\right)\right)$$
Now, $\lim_{x\to\infty}-a_x(1)=0$ and $\lim_{x\to\infty}\int_1^\infty-a_x(t)\,dt=1/2$.
This latter limit is easily computed considering the expansion
$$\arctan(z)=z-\frac{z^3}3+\frac{z^5}5-\dots$$
It follows from the squeeze theorem that
$\sum_{n\geq 1}a_x(n)=-1/2$, which complets the proof. $\square$
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