Why is it hard to reduce titanium dioxide?
Reasons why Titanium is hard to extract and purify:


The +4 oxidation state of $Ti$ is the most stable (due to its noble gas configuration) and so reducing $TiO_2$ to $Ti$ is quite difficult.
$Ti$ has chemical properties which are similar to its neighbouring elements (Vanadium, Zirconium, Chromium) and the presence of compounds containing these elements in its ores makes purification even more difficult. For example: Vanadium and Zirconium also form volatile halides ($ZrCl_4$)and oxyhalides ($VOCl_3$) which have to be separated from $TiCl_4$ by fractional distillation in Kroll's Process.




I have tried to explain why the chlorine route is taken using only thermodynamics (I'm not absolutely sure about this but these reasons are what I could come up with).

Consider these reactions which could take place in the fluidized bed reactor:


$\ce{TiO_2(s) + 2Cl_2(g) + 2C(s) -&gt; TiCl_4(g) + 2CO(g)}$
$\ce{Ti(s) + C(s) -&gt; TiC(s)}$


Comparing the Gibbs' Energies of these reactions can tell us which one is spontaneous since if the change in Gibbs' Energy is negative, the reaction will be spontaneous.
And the equation for change in Gibbs' Energy is:

$\Delta{G} = \Delta{H} - T\Delta{S}$ 

At normal temperatures, reaction 2 would occur more readily than reaction 1 since $TiC$ is extremely stable (reflected by its very high melting and boiling points) while $TiCl_4$ is unstable (it hydrolyses immediately on contact with air or oxygen to form $TiO_2$ and $HCl$). And so it is almost impossible to reduce Ti using C at low temperatures.

At higher temperatures (Kroll's Process is carried out at around $1000^oC$), $T\Delta{S}&gt;&gt;\Delta{H}$ so we can neglect $\Delta{H}$. 

For the first reaction, since all the products are gaseous (while some of the reactants are solid), $\Delta{S}$ is positive. Therefore $\Delta{G}$ will be negative and the reaction is spontaneous and will occur easily.

For the second reaction, since the number of molecules is reduced, $\Delta{S}$ is negative. Therefore $\Delta{G}$ will be positive and the reaction becomes unfavourable.



Notes: 


For the states of the reactants and products, I compared the melting and boiling points of the reactants and products with $1000^oC$ which is the temperature at which Kroll's process is carried out.
Actually, from what I read, there are many possible products for the first reaction. depending on the reaction conditions, a mixture containing $CO$, $CO_2$ and $COCl_3$ can be formed. But all these products are still gaseous so it doesn't affect the explanation.