The obvious thing to do is take the derivative with respect to $u$ of $-\frac{c}{u} \exp{(-u^2)}$ which is $c\left(2+\frac{1}{u^2}\right) \exp{(-u^2)}$.

So if $u \gt 0$ and $c$ is such that $c\left(2+\frac{1}{u^2}\right) \gt 1$ then $c\left(2+\frac{1}{x^2}\right) \gt 1$ when $x \ge u$ so $$\int_{u}^{\infty} \exp{(-x^2)} dx \lt \int_{u}^{\infty} c\left(2+\frac{1}{x^2}\right) \exp{(-x^2)} dx = \left[-\frac{c}{x} \exp{(-x^2)}\right]_{x=u}^\infty = \frac{c}{u} \exp{(-u^2)}.$$

As Chris Taylor says, $c=\frac{1}{2}$ is enough.

The obvious thing to do is take the derivative with respect to $u$ of $-\frac{c}{u} \exp{(-u^2)}$ which is $c\left(2+\frac{1}{u^2}\right) \exp{(-u^2)}$.

So if $u \gt 0$ and $c \ge \frac{1}{2}$ then $c \left(2+\frac{1}{u^2}\right) \gt 1$ and $c\left(2+\frac{1}{x^2}\right) \gt 1$ for $x \ge u$ so

$$\int_{u}^{\infty} \exp{(-x^2)} dx \lt \int_{u}^{\infty} c\left(2+\frac{1}{x^2}\right) \exp{(-x^2)} dx = \left[-\frac{c}{x} \exp{(-x^2)}\right]_{x=u}^\infty = \frac{c}{u} \exp{(-u^2)}.$$