= Solution
Take $L=S^2$. For any <Lagrangian embedding> of $S^2$ into a symplectic four-manifold, the symplectic form identifies the <normal bundle> with $T^*S^2$. The <self-intersection formula> and the <Euler characteristic> give
$$
[L]\cdot[L]
=\langle e(NL),[L]\rangle
=\langle e(T^*S^2),[S^2]\rangle
=2
$$
up to the harmless orientation sign. If a <smooth isotopy> displaced $L$, its final image would represent the same homology class but have intersection number zero with $L$, a contradiction. This is the <smooth non-displaceability from self-intersection>.
For a compact ambient example, equip $S^2\times S^2$ with $\omega\oplus(-\omega)$. Its diagonal
$$
\Delta=\{(x,x):x\in S^2\}
$$
is Lagrangian because the two summands cancel on $T\Delta$, and the preceding argument shows that it is not smoothly displaceable.
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