There is a pullback lemma which says that if a diagram is commutative and the right square is a pullback then the left square is a pullback if and only if the outer rectangle is a pullback.
So, are there any simple counterexamples where the outer rectangle is a pullback but left and right squres are not?
Consider the following diagram, where $1$ is a terminal object, $X$ is any nonterminal object, and $f$ is any arrow $1\to X$. The other arrows are forced to be the unique arrows to $1$, and the squares automatically commute.
$$ \require{AMScd} \begin{CD} 1 @>{f}>> X @>>> 1\\ @VVV @VVV @VVV\\ 1 @>>> 1 @>>> 1 \end{CD} $$
Since the pullback of any arrow along an identity map is the same arrow, the interior squares are not pullbacks, but the outer rectangle is.
In more detail, it's clear that these squares are pullbacks:
$$ \require{AMScd} \begin{CD} X @>{\mathrm{id}_X}>> X\\ @VVV @VVV \\ 1 @>>> 1 \end{CD} \hspace{.5in} \begin{CD} 1 @>>> 1\\ @VVV @VVV \\ 1 @>>> 1 \end{CD} $$
... and pullbacks (like all limits) are unique up to unique isomorphism, so the interior squares aren't pullbacks unless $X\cong 1$. But we've assumed $X$ is not terminal.