You can prove this using diagonalization. Let $R_1,R_2,\ldots$ be an enumeration of the infinite regular languages. We will construct a descending sequence $L = L_0 \supseteq L_1 \supseteq L_2 \supseteq \cdots$ and an increasing sequence $X_0 = \emptyset \subseteq X_1 \subseteq X_2 \subseteq \cdots$ such that $R_n \setminus L_n \neq \emptyset$, and furthermore $L \setminus L_n$ and $X_n$ are finite for all $n$, $X' = \bigcup_n X_n$ and $L' = \bigcap_n L_n$ are infinite, and $L' \supseteq X'$.
At stage $t$, we construct $L_t$ and $X_t$. If $R_t \setminus L_{t-1} \neq \emptyset$, we set $L_t = L_{t-1}$. Otherwise, $R_t \subseteq L_{t-1}$. The set $\Delta = L_{t-1} \setminus R_t$ is infinite, since otherwise $L = R_t \cup \Delta \cup (L \setminus L_{t-1})$ would be regular, and in particular we can choose some $x \in \Delta \setminus X_{t-1}$ and set $L_t = L_{t-1} - x$. In both cases we set $X_t = X_{t-1} + \min L_{t-1}$.
By construction $R_n \setminus L' \supseteq R_n \setminus L_n \neq \emptyset$, and so $L'$ is not equal to any infinite regular language, and so to no regular language.
This arguments works if we replace the class $\mathcal{L}$ of regular languages with any other class which is closed under word addition (i.e., $L \in \mathcal{L}$ implies $L+w \in \mathcal{L}$). The argument is constructive in the sense that the $t$th smallest word is either in $X_t$ or in $L\setminus L_t$. When $L$ is itself regular and $\mathcal{L}$ is the class of regular languages, this makes the resulting language decidable, thus improving on the cardinality argument given by Nathan Dunn. While in this particular case there are more direct solutions, diagonalization could work in more general situations.