Some thoughts.
We rephrase the problem as follows.
Problem. Let $u \in (0, 1)$ be given. Let
$$f(x) := \frac{x}{u}\left(\frac{1}{\sin x - u}-\frac{1}{\sin x + u}\right)-\frac{1}{x-u}-\frac{1}{x+u}$$
for $x\in (\arcsin{u}, \pi/2)$.
Prove that $f$ is convex on $(\arcsin{u}, \pi/2)$.
Denote $c := \cos x, s := \sin x$. We have
\begin{align*}
f''(x) &= \frac{2xc^2}{u}\left(\frac{1}{(s-u)^3} - \frac{1}{(s+u)^3}\right)\\
&\quad
+ \frac{xs - 2c}{u}\left(\frac{1}{(s-u)^2} - \frac{1}{(s+u)^2}\right) - \frac{2}{(x-u)^3} - \frac{2}{(x+u)^3}.
\end{align*}
It is easy to prove that $c \le 1 - \frac12 sx$ for all $x\in [0, \pi/2]$. We have
\begin{align*}
f''(x) &\ge \frac{2xc^2}{u}\left(\frac{1}{(s-u)^3} - \frac{1}{(s+u)^3}\right)\\
&\qquad
+ \frac{xs - 2\cdot (1 - \frac12 sx) }{u}\left(\frac{1}{(s-u)^2} - \frac{1}{(s+u)^2}\right)\\
&\qquad - \frac{2}{(x-u)^3} - \frac{2}{(x+u)^3}.\tag{1}
\end{align*}
From (1), to prove that $f''(x) \ge 0$, after clearing the denominators, it suffices to prove that
$$q_4 v^4 + q_3 v^3 + q_2 v^2 + q_1v + q_0 \ge 0 \tag{2}$$
where $v := u^2$ and
\begin{align*}
q_4 &:= 3\,{s}^{2}x-2\,s+2\,x , \\
q_3 &:= {s}^{4}x-9\,{s}^{2}{x}^{3}+2\,{s}^{3}-12\,{s}^{2}x+6\,s{x}^{2}+4\,{x}^
{3}, \\
q_2 &:= -3\,{s}^{4}{x}^{3}+9\,{s}^{2}{x}^{5}+9\,{s}^{4}x-6\,{s}^{3}{x}^{2}+6\,
{s}^{2}{x}^{3}-6\,s{x}^{4}-3\,{x}^{5},\\
q_1 &:= 3\,{s}^{4}{x}^{5}-3\,{s}^{2}{x}^{7}-3\,{s}^{6}x+3\,{s}^{4}{x}^{3}+6\,{
s}^{3}{x}^{4}-9\,{s}^{2}{x}^{5}+2\,s{x}^{6}+{x}^{7}, \\
q_0 &:= -{s}^{4}{x}^{7}-{s}^{6}{x}^{3}-2\,{s}^{3}{x}^{6}+3\,{s}^{2}{x}^{7}.
\end{align*}
(2) is true for all $x\in [0, \pi/2]$ and $v \in [0, 1]$ which is verified (in several seconds) by Mathematica - a Computer Algebra System (CAS). A human verifiable proof is required.
About the verifying:
We use Mathematica to verify a stronger inequality:
$$p_4 v^4 + p_3 v^3 + p_2 v^2 + p_1v + p_0 \ge 0$$
where $v \in [0, 1]$ and $x \in [0, 2]$ and $0 \le w \le 1$ and
$$x - x^3/6 \le w \le x - x^3/6 + \frac{x^5}{120},$$
and
\begin{align*}
p_4 &:= 3\,{w}^{2}x-2\,w+2\,x , \\
p_3 &:= {w}^{4}x-9\,{w}^{2}{x}^{3}+2\,{w}^{3}-12\,{w}^{2}x+6\,w{x}^{2}+4\,{x}^
{3}, \\
p_2 &:= -3\,{w}^{4}{x}^{3}+9\,{w}^{2}{x}^{5}+9\,{w}^{4}x-6\,{w}^{3}{x}^{2}+6\,
{w}^{2}{x}^{3}-6\,w{x}^{4}-3\,{x}^{5},\\
p_1 &:= 3\,{w}^{4}{x}^{5}-3\,{w}^{2}{x}^{7}-3\,{w}^{6}x+3\,{w}^{4}{x}^{3}+6\,{
w}^{3}{x}^{4}-9\,{w}^{2}{x}^{5}+2\,w{x}^{6}+{x}^{7}, \\
p_0 &:= -{w}^{4}{x}^{7}-{w}^{6}{x}^{3}-2\,{w}^{3}{x}^{6}+3\,{w}^{2}{x}^{7}.
\end{align*}
(Note: We use $x - x^3/6 \le w \le x - x^3/6 + \frac{x^5}{120}$ to avoid dealing with $\sin x$.)
f[x_, w_,
v_] := (3w^2x - 2w + 2x)*
v^4 + (w^4x - 9w^2x^3 + 2w^3 - 12w^2x + 6wx^2 + 4x^3)
v^3 + (-3w^4x^3 + 9w^2x^5 + 9w^4x - 6w^3x^2 + 6w^2x^3 -
6wx^4 - 3x^5)
v^2 + (3w^4x^5 - 3w^2x^7 - 3w^6x + 3w^4x^3 + 6w^3x^4 -
9w^2x^5 + 2wx^6 + x^7)v - w^4x^7 - w^6x^3 - 2w^3x^6 +
3w^2*x^7
pos1 := v >= 0 && v <= 1 && w >= x - x^3/6 &&
w <= x - x^3/6 + x^5/120 && x >= 0 && x <= 2 && w <= 1
Resolve[ForAll[{x, w, v}, pos1, f[x, w, v] >= 0], Reals]
