Let $B_{t} \sim N(0,t)$, and show that $E[\lvert B_{t}\rvert ] < \infty$
My idea:
$E[\lvert B_{t}\rvert ]=\frac{1}{\sqrt{2\pi}t}\int_{-\infty}^{\infty}\lvert y\rvert e^{-\frac{y^2}{2t}}dy$
and then by symmetry of the probability density function, we have
$\int_{-\infty}^{\infty}\lvert y\rvert e^{-\frac{y^2}{2t}}dy=2\int_{0}^{\infty}ye^{-\frac{y^2}{2t}}dy$
Now note that $\exp(y)\geq y$ on $\mathbb R$. So we obtain the bound
$\int_{0}^{\infty}ye^{-\frac{y^2}{2t}}dy\leq \int_{0}^{\infty}\exp(y-\frac{y^2}{2t})dy$
By completing the square: $y -\frac{y^{2}}{2t}=-\frac{(y^2-2ty+t^2-t^2)}{2t}=-\frac{(y-t)^2-t^2}{2t}$ So:
$\int_{0}^{\infty}\exp(y-\frac{y^2}{2t})=\int_{0}^{\infty}\exp(-\left(\frac{(y-t)^2-t^2}{2t}\right))dy=\exp(\frac{t}{2})\int_{0}^{\infty}\exp(-\left(\frac{y-t}{\sqrt{2t}}\right)^{2})dy$
And we set $z = \frac{y-t}{\sqrt{2t}}$ to deliver
$\int_{0}^{\infty}\exp(\left(\frac{y-t}{\sqrt{2t}}\right)^{2})dy\leq \int_{-\infty}^{\infty}\exp(-z^{2})dz = \sqrt{\pi}$
Does this suffice?
You can either use Jensen inequality as stated in the comment by user Kavi Rama Murthy or find it directly.
First for each, $B_t$ has the same distribution as $\sigma B_1 =\sqrt t B_1$ and $B_1$ is standard normal random variable. Now you calculate
$$\mathbb E|\sigma B_1|=\mathbb E \sigma B_1 -2 \sigma\mathbb E( B_1 : B_1\leq 0)$$ And the last integral is simply $$\int_{-\infty}^0 x\phi(x) dx=-\int_0^\infty x\phi(x) dx=-\frac{1}{\sqrt{2\pi}}\int_0^\infty e^{-u}du=-\frac{1}{\sqrt{2\pi}}$$ which gives $$\mathbb E |B_t|=-2\sigma \cdot-\frac{1}{\sqrt{2\pi}}=\sqrt{\frac{2t}{\pi}}$$