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For every $l\ge 1$, take a plane immersed circle with $l$ transverse double points and form its product with an embedded circle in a second symplectic plane. The resulting torus is a Product Lagrangian immersion with $l$ clean double circles. Resolving them gives a connected nonorientable Lagrangian embedding in ${\mathbb{R}}^4$ with Euler characteristic $-4l$.

Take $L=S^2$. For any Lagrangian embedding of $S^2$ into a symplectic four-manifold, the symplectic form identifies the normal bundle with $T^{\ast }{S}^2$. The self-intersection formula and the Euler characteristic giveup 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 diagonalis Lagrangian because the two summands cancel on $T\Delta$, and the preceding argument shows that it is not smoothly displaceable.