Context-free languages are not closed under intersection

The grammar $S_1\to AB$, $A\to aAb\mid \epsilon$, $B\to cB\mid \epsilon$ generates $L_1$, and the grammar $S_2\to CD$, $C\to aC\mid \epsilon$, $D\to bDc\mid \epsilon$ generates $L_2$. Therefore both languages are context-free by [L1].

A word belongs to $L_1\cap L_2$ exactly when its number of $a$'s equals its number of $b$'s and its number of $b$'s equals its number of $c$'s. Hence $L_1\cap L_2=\{a^n{b}^n{c}^n:n\ge 0\}$.

Suppose $L_1\cap L_2$ were context-free. Let $p$ be the pumping length from [L2], and consider the word $a^p{b}^p{c}^p$. Any factor $vwx$ with $\mathrm{\mid }vwx\mathrm{\mid }\le p$ lies within one block or crosses only one of the two block boundaries. Pumping $v$ and $x$ therefore changes at most two of the three symbol counts, so for $i=0$ or $i=2$ the word $u{v}^{i}w{x}^{i}y$ cannot still have equal numbers of $a$'s, $b$'s, and $c$'s. This contradicts [L2].

Thus $L_1\cap L_2$ is not context-free even though each of $L_1$ and $L_2$ is. So context-free languages are not closed under intersection.