$$f(t) = \frac{\sqrt{1+t} - \sqrt{|1-t|}}{1+\sqrt{t}}$$
and observe that $f(1/t) = f(t)$, and that $f(0) = 0, f(1) = 1/\sqrt{2}$. Therefore, it suffices to prove that $2(f(s^2))^2 \le 1$ if $0 \le s \le 1$, i.e, that
$$2(1 + s^2 + 1 - s^2 - 2\sqrt{1 - s^4}) \le (1 + s)^2.$$
This is equivalent to $4-(1+s)^2 \le 4\sqrt{1-s^4}$, or $(4-(1+s)^2)^2 \le 16(1-s^4)$, i.e.
$$(1-s)^2(3+s)^2 \le 16(1-s)(1+s+s^2+s^3).$$
Since $0 \le s \le 1$ this holds provided that
$$(1-s)(9+6s+s^2) \le 16(1+s+s^2+s^3),$$
equivalently, iff
$$9 - 3s - 5s^2 - s^3 \le 16(1 + s + s^2 + s^3),$$
which clearly holds for all $s \ge 0$.

We first show that
$$\sqrt{a} - \sqrt{b} \le \sqrt{a-b} \quad \text{if} \quad 0 \le b \le a. \qquad (9)$$
Indeed, we have $0 \le \sqrt{b} \le \sqrt{a}$ and multiply by $2\sqrt{b}$ to get $0 \le 2b \le 2\sqrt{ab}$.
If we add $a - b - 2\sqrt{ab}$ to both sides this yields
$$(\sqrt{a} - \sqrt{b})^2 = a - 2\sqrt{ab} + b \le a - b.$$
Because $a \ge b$, we can take square roots to obtain $\sqrt{a} - \sqrt{b} \le \sqrt{a-b}$.
Next, we note that the inequality we wish to show is unchanged when $t$ is
replaced by $1/t$. Hence, it suffices to consider $0 \le t \le 1$. In this case,
$|1-t| = 1-t$. With $a = 1+t$ and $b = 1-t$ we obtain from (9) the inequality
$$\sqrt{1+t} - \sqrt{1-t} \le \sqrt{2t} = \frac{2\sqrt{t}}{\sqrt{2}} \le \frac{1+\sqrt{t}}{\sqrt{2}},$$
where we have used $\sqrt{t} \le 1$ in the last step. This is the desired result.