The answer is yes.

A topological space has *calibre* $\aleph_1$ if for every uncountable sequence $\langle U_\alpha\mid\alpha\lt\aleph_1\rangle$ of nonempty open sets $U_\alpha\subset X$, there is an uncountable subfamily $\Lambda\subset\aleph_1$ with $\bigcap_{\alpha\in\Lambda}U_\alpha\neq\emptyset$. 

<b>Theorem.</b> If a topological space $X$ has calibre $\aleph_1$, then so does $X\times X$.

Proof. Suppose $X$ has calibre $\aleph_1$, and consider a family of nonempty open sets $U_\alpha$ in $X\times X$, for $\alpha\lt\aleph_1$. The idea will be to reduce first on the first coordinate, and then on the second. Since $U_\alpha$ is open in the product, we may choose a nonempty open rectangle $U_\alpha^0\times U_\alpha^1\subset U_\alpha$. Consider the sequence $\langle U_\alpha^0\mid\alpha\lt\aleph_1\rangle$ of open sets arising from the first coordinate. By the calibre $\aleph_1$ property of $X$, it follows that there is $\Lambda_0\subset\aleph_1$ with some $x\in\bigcap_{\alpha\in\Lambda_0}U_\alpha^0$. Consider now the enumeration $\langle U_\alpha^1\mid \alpha\in\Lambda_0\rangle$. By the calibre property again, we may find a smaller uncountable family $\Lambda\subset\Lambda_0$ with some $y\in\bigcap_{\alpha\in\Lambda}U_\alpha^1$. It follows that $(x,y)\in\bigcap_{\alpha\in\Lambda}U_\alpha^0\times U_\alpha^1\subset\bigcap_{\alpha\in\Lambda}U_\alpha$, and so $X\times X$ has calibre $\aleph_1$, as desired. QED