Let enumerate the people $1,2,\cdots,n $. Let denote the house which is assigned to the man numbered $n$ as $n$. In this way $n$-th house is assigned to $n$-th man. List of preferencies of $i$-th man can be represent by a permutation $\sigma_{i}$. Now, in oreder to prove the assertion, let assume on the contrary that $\sigma_i(1) \neq i,\, i=1,2,\cdots,n$. The idea is to construct a permutation $\sigma$ (not equal to id) for which $\sigma(i) = i $ or $\sigma(i) = \sigma_i(1)$. If such a permutation exists then it will be a contradiction with the problem clause. Let $a_1 = 1, \, b_1 = \sigma_1(1);\, a_{i+1} = b_i,\, b_{i+1} = \sigma_{a_{i+1}}(1)$. Note that $b_i \neq a_i$. In this way there will be a moment when for the first time $b_k = b_s,\, s < k$. Then we define the permutation $\sigma$ as follows: If $ i=a_j $ for some $ j=s+1,s+2,\cdots, k $ then $\sigma(i) = b_j $ else $\sigma(i) = i$.

More dirty-hands type of approach: Denote the house assigned to $i^{th}$ person as $a_i$. Start with an arbitrary person $i_1$, who did not receive his top choice. Suppose that, his top choice is $a_{i_2}$ (namely, with person $i_2$). Switching these two houses makes the first person better off, hence, in $i_2$'s list, $a_{i_2}>a_{i_1}$, otherwise would contradict with the preamble. In particular, $i_2$'s top choice is not $a_{i_1}$. Now, if $i_2$ has his top house, we are done. Otherwise, suppose, $a_{i_3}$ is his top choice. We make the following observations: Claim: $i_3$'s top choice is neither $a_{i_1}$, nor $a_{i_2}$. Proof:Clearly, it is not $a_{i_2}$, since, swapping $a_{i_2}$ and $a_{i_3}$ makes $i_2$ better off, it must make $i_3$ worse, hence, in $i_3$'s list, $a_{i_3}>a_{i_2}$. To see why it cannot be $a_{i_1}$ either, notice that had it been, then, the following assignment, $$ (i_1 \mapsto a_{i_2}, i_2 \mapsto a_{i_3}, i_3 \mapsto a_{i_1}), $$gives us everyone's top choice, a contradiction. Hence, $i_3$'s top choice is some other $i_4$. Claim: $i_4$'s top choice is neither $a_{i_1}$, nor $a_{i_2}$ or $a_{i_3}$. Proof: This is a mere repetition of the previous argument. For the sake of completeness, it is included. First, it cannot be $i_3$ (follows from swapping $i_3$ and $i_4$). Had it been $i_2$, then, $$ (i_2\mapsto a_{i_3},i_3 \mapsto a_{i_4},i_4 \mapsto a_{i_2}), $$would assign at least three people to their top choices, a contradiction. Finally, to see why it cannot be $a_{i_1}$, notice that, this time, we need, $$ (i_1\mapsto a_{i_2},i_2\mapsto a_{i_3},i_3\mapsto a_{i_4},i_4\mapsto a_{i_1}), $$again, contradicts with the statement. Hence, continuing in this manner, at each step, $i_k$'s top choice is none of $a_{i_j},j=1,2,\dots,k-1$. Hence, we must stop at somewhere, $i_n$, and for this person, his top choice is none of $a_{i_1},a_{i_2},\dots,a_{i_{n-1}}$, thus, he must be assigned to his top choice.