Prove $$\sqrt{1-e^{-1}}<\frac{1}{\sqrt{\pi}}\int^1_0 e^{-x^2}dx<\sqrt{1-e^{-2}}.$$
Since $$ \forall x\in[0,1],\quad e^{-x^2}=\sum_{n\geq 0}\frac{(-1)^n}{n!}\,x^{2n}$$ and by integrating termwise over $[0,1]$ we get: $$ \int_{0}^{1}e^{-x^2}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)\,n!}.$$ This will help?
Assuming you have a multiplication by 2 in the integral, otherwise the lower bound does not hold (as @PierreCarre noted).
Squaring, we want to prove $$ 1-e^{-1}<\frac4\pi\iint_{[0,1]\times[0,1]}e^{-x^2-y^2}\,\mathrm{d}x\,\mathrm{d}y<1-e^{-2} $$ Noting that the unit square $[0,1]^2$ is squeezed(*) between the quarter discs of radius 1 and radius $\sqrt{2}$, and $$ \int_0^a e^{-r^2}r\,\mathrm{d}r=\frac{1-e^{-a^2}}{2} \quad \text{and}\quad \int_0^{\pi/2} \frac4\pi\,\mathrm{d}\theta=2 $$ give our two bounds.
(*): We have for all $(x,y)\in[0,1]^2$, that $r^2=x^2+y^2\leq 2$ and $\theta\in[0,\frac\pi2]$. On the other hand, for $r\leq 1$ and $\theta\in[0,\frac\pi2]$, we have $x=r\cos\theta\in[0,1]$ and $y=r\sin\theta\in[0,1]$. So $$ (0\leq r\leq 1,0\leq\theta\leq\frac\pi2)\subseteq[0,1]^2\subseteq (0\leq r\leq\sqrt2,0\leq\theta\leq\frac\pi2). $$
