You're right, it is the fact that we are considering the disjoint union topology that makes it non-second-countable. While Asaf's proof is very slick, here's a (much longer) direct argument.
Let $A$ be an uncountable set, and let $X_\alpha=\mathbb{R}$ for each $\alpha\in A$. Let $X=\coprod_{\alpha\in A}X_\alpha$, with the disjoint union topology. The open sets of $X$ are those sets of the form $\coprod_{\alpha\in A}U_\alpha$ where $U_\alpha\subseteq X_\alpha$ is open. In particular, for any $\beta\in A$, $X_\beta=\coprod_{\alpha\in A}U_\alpha$ where $U_\beta=X_\beta$ and $U_\alpha=\emptyset$ for $\alpha\neq\beta$ is an open set of $X$.
Let $\mathcal{B}$ be any basis for the topology on $X$, so that every open set $U\subseteq X$ has $U=\bigcup_{\substack{B\,\in\,\mathcal{B}\\B\,\subseteq\, U}} B.$ Obviously, this implies that for any non-empty open set $U$ in $X$, there exists a non-empty $B\in\cal{B}$ such that $B\subseteq U$. In other words, for any non-empty open set $U$ in $X$, we have $|Z(U)|\geq 1$, where $Z(U)=\{B\in\mathcal{B}\mid B\subseteq U,B\neq\emptyset\}.$ Because each $X_\alpha$ is a non-empty open set in $X$, and $X_\alpha\cap X_\beta=\emptyset$ for any $\alpha\neq\beta$, we have that $Z(X_\alpha)\cap Z(X_\beta)= \{B\in\mathcal{B}\mid B\subseteq X_\alpha\cap X_\beta,B\neq\emptyset\}=\emptyset\text{ for any }\alpha\neq\beta.$
Because $\mathcal{B}\supseteq\bigcup_{\alpha\in A}Z(X_\alpha)$ and $Z(X_\alpha)\cap Z(X_\beta)=\emptyset$ for $\alpha\neq\beta$, we have $|\mathcal{B}|\geq\sum_{\alpha\in A}|Z(U)|\geq\sum_{\alpha\in A}1=|A|,$ Because $A$ is uncountable, $\mathcal{B}$ is uncountable. Thus, $X$ is not second-countable.