Preparation of alkyl halide from alcohols
It can happen.

It practice, you avoid it happening by keeping the temperature low. $\ce{Br-}$ is a better nucleophile than $\ce{H2O}$ is a base, so at lower temperatures, the substitution is faster. 

If the elimination happens, alkenes react with $\ce{HBr}$ to form alkyl bromides. As long as acid is present, the carbocation can reform. For example, with isopropyl alcohol

Direct SN1 Substitution 
$$\ce{(CH3)2CH-OH + HBr -&gt; (CH3)2CH-OH2+ + Br-}$$
$$\ce{(CH3)2CH-OH2+ -&gt; (CH3)2CH+ + H2O}$$
$$\ce{(CH3)2CH+ + Br- -&gt; (CH3)2CH-Br}$$

Substitution by elimination-addition
$$\ce{(CH3)2CH-OH + HBr -&gt; (CH3)2CH-OH2+ + Br-}$$
$$\ce{(CH3)2CH-OH2+ -&gt; (CH3)2CH+ + H2O}$$
$$\ce{(CH3)2CH+ + H2O &lt;=&gt; CH3-CH=CH2 + H3O+}$$
$$\ce{(CH3)2CH+ + Br- -&gt; (CH3)2CH-Br}$$

In some cases, two processes may appear to generate different alkyl halides, but this would be no different than if a carbocation rearrangement occurred.