Since the real line is a linearly ordered topological space, open sets may have been defined as those sets that are unions of open intervals, i.e., unions of sets of the form $(a,b)$. In that case suppose that $U_0$ and $U_1$ are open sets. Then there are index sets $A_0$ and $A_1$ and families $\mathscr{I}_0=\{I_{0,\alpha}:\alpha\in A_0\}$ and $\mathscr{I}_1=\{I_{1,\alpha}:\alpha\in A_1\}$ of open intervals such that $U_0=\bigcup_{\alpha\in A_0}I_{0,\alpha}\qquad\text{and}\qquad U_1=\bigcup_{\alpha\in A_1}I_{1,\alpha}\;.$
Then
$\begin{align*} U_0\cap U_1&=\left(\bigcup_{\alpha\in A_0}I_{0,\alpha}\right)\cap\left(\bigcup_{\beta\in A_1}I_{1,\beta}\right)\\ &=\bigcup_{\alpha\in A_0}\left(I_{0,\alpha}\cap\bigcup_{\beta\in A_1}I_{1,\beta}\right)\\ &=\bigcup_{\alpha\in A_0}\left(\bigcup_{\beta\in A_1}\Big(I_{0,\alpha}\cap I_{1,\beta}\Big)\right)\\ &=\bigcup_{\langle\alpha,\beta\rangle\in A_0\times A_1}\Big(I_{0,\alpha}\cap I_{1,\beta}\Big)\;.\tag{1} \end{align*}$
The intersection of two open intervals is an open interval (possibly empty), so $(1)$ is a union of open intervals and as such is an open set.
This shows that the intersection of two open sets is open, and the result for any finite collection of open sets is easily established by induction. Suppose that you know that the intersection of $n$ open sets is always open. Let $U_1,\dots,U_{n+1}$ be any $n+1$ open sets. Let $U=U_1\cap\ldots\cap U_n$; by the induction hypothesis $U$ is open. Then $U_1\cap U_2\cap\ldots\cap U_n\cap U_{n+1}=U\cap U_{n+1}$, which, being the intersection of two open sets, is open by the result proved above.