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Let $f(x)=ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}$.

Assume $f(x_1)=f(x_2)$. We have then

\begin{gather*} a\left(x_1+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a} = a\left(x_2+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a} \iff\\ \left(x_1+\frac{b}{2a}\right)^2 = \left(x_2+\frac{b}{2a}\right)^2 \iff\\ x_1+\frac{b}{2a} = x_2+\frac{b}{2a} \vee x_1+\frac{b}{2a} = -x_2-\frac{b}{2a}\iff\\ x_1 = x_2\vee x_1 = -x_2-\frac{b}{a} \end{gather*}

so $f$ is not injective.

How would I partition $\mathbb{R}$ into two intervals where $f$ is injective using this?

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A quadratic is injective on the intervals $(-\infty, -b/2a]$ and $[-b/2a,\infty)$ as well as any interval properly contained in one of those. There are other regions, though not other intervals.

To see why, examine the quadratic equation. $f(x)=k$ has solutions exactly where $ax^2+bx+(c-k)$ has roots. You need to capture either the $+$ or the $-$ but not both for the function to be injective, which means staying on one side of $-b/2a$.

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    @user159517 the empty interval is properly contained in these intervals. I'm not saying these two are the only ones, I'm saying these two and all of their proper subintervals are – 2017-01-22
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    I know this to be true but I would quite like an argument as to why those two intervals do the job. – 2017-01-22
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    @Neutronic I said why. Because by the quadratic equation the interval can only contain numbers on one side of $-b/2a$ – 2017-01-22
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    How does this follow from the quadratic equation? – 2017-01-22
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    @Neutronic Please point to the words in my second proof that doesn't make sense to you, and why you think it's not true/might not be true. – 2017-01-22