(Heuristic) In general, we expect there is an $(n+1)$-dimensional rectifiable current $\mathcal{C}$ with $\partial\mathcal{C}=L-L^{\prime}$ constructed from universal families of holomorphic curves with boundary on $L$ and $L^{\prime}$. The holomorphic curves $u:\Sigma\to X$ can appear in three types:

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  Automatically transverse holomorphic curves: there exist first order deformations $v_{1},\ldots,v_{n-1}$ such that $dF=\Omega(\cdot,v_{1},\ldots v_{n-1})$ does not vanish identically as a 1-form on $\Sigma$ (cf. section 3.3).

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  Nonconstant holomorphic curves, which are not automatically transverse. We expect their boundary evaluation to be contained in a Hausdorff dimension $\leq n-1$ subset of $\text{supp}(L)\cup\text{supp}(L^{\prime})$ (cf. section 3.3).

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  Constant holomorphic maps $u:\Sigma\to\text{supp}(L)\cap\text{supp}(L^{\prime})$. These would only arise if $L$ and $L^{\prime}$ have some overlapping support, so did not appear in our previous discussions. For dimensional reasons, these cannot contribute to the $(n+1)$-dimensional current $\mathcal{C}$.

  The key difference from the second case is that at interior points of $\text{supp}(L)\cap\text{supp}(L^{\prime})$, there are $n$ linearly independent first order deformations, such that $v_{1},\ldots v_{n}$ span $TL$ upon boundary evaluation. This behaviour can only be compatible with $dF=0$ for constant curves.

We now impose the special Lagrangian condition, and consider the automatically transverse case. Along $\partial\Sigma$, the counterclockwise directional derivative of $F$ has argument equal to the constant Lagrangian angle $\hat{\theta}$ modulo $\pi\mathbb{Z}$. As such we expect $F(\partial\Sigma)$ to be contained in a line segment with incline angle $\hat{\theta}$. By the maximum principle on the holomorphic function $F$, the entire $F(\Sigma)\subset\mathbb{C}$ is contained in a line segment. However, the open mapping theorem in complex analysis then implies $F$ is constant, which rules out the automatically transverse curves.

Now the only contributions to $\mathcal{C}$ would come from the nonconstant, not automatically transverse curves. This forces $\text{supp}(\partial\mathcal{C})\cap(\text{supp}(L)\cup\text{supp}(L^{\prime}))$ to be contained in a Hausdorff $(n-1)$-dimensional subset. However $\partial\mathcal{C}=L-L^{\prime}$ as integral currents, so the $n$-dimensional current $L-L^{\prime}$ has support dimension $\leq n-1$, which forces it to vanish. This shows $L=L^{\prime}$. ∎