Since $x > 0$, your inequality can be re-stated as $f(x) = 1+x \leq e^{\sqrt{x}} = g(x)$. Observe that $f(0) = g(0) = 1$. Hence, it is enough to show that $f'(x) < g'(x)$ for all $x > 0$. That is,
$1 \leq \frac{1}{2\sqrt{x}}e^{\sqrt{x}}$
Calling $\sqrt{x} = y$,
$a(y) = 2y \leq e^{y} = b(y)$
Now, for $y \leq 1$, observe that $e^{y} \geq 1+y \geq 2y$. For $y > 1$, it is enough to show that $a'(y) \leq b'(y)$, that is $2 \leq e^{y}$. Since $y > 1$, $e^{y} > e > 2$, which completes the proof.
Back to the main question, recall that:
$E[\log(1+0.5X)^{2}] = E[I(X \geq 0)\log(1+0.5X)^{2}] + E[I(-1 \leq X \leq 0)\log(1+0.5X)^{2}]$
Hence, it is enough to show that both terms on the RHS are finite. Using your inequality, the following holds with probability $1$,
$0 \leq I(X \geq 0)\log(1+0.5X)^{2} \leq I(X \geq 0)0.5X$
and
$E[I(X \geq 0)\log(1+0.5X)^{2}] \leq E[I(X \geq 0)0.5X] < \infty$
Since $I(-1 \leq X \leq 0)\log(1+0.5X)^{2}$ is bounded, the proof is complete.