The others have explained the decay process of radioactive material very well. Therefore, I'm not going to elaborate the same thing again, but want to point out certain thing you seemingly do not understand clearly. In your question, you states that:
I had previously thought that the definition of a half-life is the time it takes for the amount of material to half in its decay process.
That statement is not quite to the point. The decay doesn't mean it vanish (or disappear) to air. It is not mass decay (kind of, theoretically but some mass remains, e.g., as $\ce{^{206}Pb}$, which is stable and not radioactive). The process is complicated one. For instance, see the total decay process for $\ce{^{238}_{92}U -> ^{206}_{82}Pb}$:
$$\ce{^{238}U ->[t_{1/2} = 4.4 \cdot 10^9 y] ^{234}Th ->[t_{1/2} = 24.1 d] ^{234}Pa ->[t_{1/2} = 46.69 h] ^{234}U ->[t_{1/2} = 2.455 \cdot 10^5 y] ^{230}Th \\ ->[t_{1/2} = 7.54 \cdot 10^4 y] ^{226}Ra ->[t_{1/2} = 1599 y] ^{222}Rn ->[t_{1/2} = 3.82 d] ^{218}Po ->[t_{1/2} = 3.04 min] ^{214}Pb ->[t_{1/2} = 27 min] ^{214}Bi\\ ->[t_{1/2} = 19.9 min] ^{210}Po ->[t_{1/2} = 160 \mu s] ^{206}Pb}$$
Therefore, for novices, what half-life simply means is the original radioactivity of given material to become half of its initial value (Refer to TAR86's answer). Thus, I decide to explain this process using your graph:
Radioactive decay of any active material is a spontaneous process, which follows first order kinetics:
$$\alpha = \alpha_\circ e^{-\beta t} \tag{1}$$
where $\alpha$ is activity of the material at any time $t$ and $\alpha_\circ$ is activity of the material at time you started to measure, $t=0$. The constant $\beta$ depends on several factors including decay process (e.g., $\beta$ is not same for $\ce{U}$ and $\ce{Po}$). We can simplify this as:
$$\frac{\alpha}{\alpha_\circ } = e^{-\beta t} \Rightarrow \ln \left(\frac{\alpha}{\alpha_\circ }\right) = -\beta t \Rightarrow \ln \alpha = \ln \alpha_\circ -\beta t \tag{2}$$
This is an equation for straight line, the slope of which is equal to $\beta$ and $y$-interception is $\ln \alpha_\circ$. By definition, $t_{1/2}$ is the time when $\alpha = \frac{1}{2} \alpha_\circ$. Applying this to equation $(2)$ gives:
$$\ln \frac{\alpha_\circ}{2} = \ln \alpha_\circ -\beta t_{1/2} \quad \Rightarrow \quad \therefore \; t_{1/2} = \frac{\ln 2}{\beta} \tag{3}$$
Thus, you can find $t_{1/2}$ by just getting $\beta$ from above straight line (Note that $t_{1/2}$ is independent of $\alpha_\circ$). Unfortunately, you don't have that straight line here. But still, you can find $t_{1/2}$ by analysing the given graph.
The equation of your graph is equation $(1)$. According to your graph, at $t=0$, activity has been measured as $\pu{16000 decays/min}$, which is your $\alpha_\circ$. Thus, $\frac{1}{2} \alpha_\circ$ should be $\pu{8000 decays/min}$ (see above graph). Accordingly, time taken to decay $\pu{16000 decays/min \rightarrow 8000 decays/min}$ is apparently $\pu{8 d}$. Therefore, $t_{1/2}$ is $\pu{8 d}$. If you uncertain of the value, you can check the next half-time by finding how much time it takes to decay $\pu{8000 decays/min \rightarrow 4000 decays/min}$. Not surprisingly, it is also $\pu{8 d}$ and so on (Note: If you choose $\alpha_\circ = \pu{12000 decays/min}$, you'd see time taken to decay $\pu{12000 decays/min \rightarrow 6000 decays/min}$ is still $\pu{8 d}$).
To go to an extra mile, now you can calculate the constant $\beta$ for this process. From the eqtation $(2)$:
$$\beta = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{\pu{8 d}} = \pu{0.087 d-1}$$
