Simple highest-weight modules are classified by highest weight

Assume $V$ and $W$ have the same highest weight $\lambda$; then $\lambda$ is dominant integral by [L3], so $L(\lambda )$ exists and is finite dimensional by [L4].

Choose a highest weight vector $v_{\lambda }$ of $V$; by [L5] the classification relations hold in $V$, so the map $\phi :M_{\mathrm{i}\mathrm{n}\mathrm{t}}(\lambda )\to V$ with $\phi (u+I_{\lambda })=u\cdot v_{\lambda }$ is a well-defined surjective module map; its kernel $K$ is a submodule of $M_{\mathrm{i}\mathrm{n}\mathrm{t}}(\lambda )$ and is proper because $v_{\lambda }\mathrm{\ne }0$.

Conversely an isomorphism $\psi :V\to W$ of $\mathfrak{g}$-modules carries weights to weights bijectively (it intertwines the $\mathfrak{h}$-action), so the set of weights of $W$ is the image of that of $V$; by [L1] and [L2] each of $V$ and $W$ has a unique highest weight, and uniqueness of the maximum of a finite weight set under the partial order gives $\lambda (V)=\lambda (W)$.

The quotient $M_{\mathrm{i}\mathrm{n}\mathrm{t}}(\lambda )\mathrm{/}K\cong V$ is simple by hypothesis, so $K$ is a maximal proper submodule of $M_{\mathrm{i}\mathrm{n}\mathrm{t}}(\lambda )$; by [L4] the unique maximal proper submodule is $N_{\lambda }$, so $K=N_{\lambda }$ and $V\cong L(\lambda )$.

Hence any two finite-dimensional simple highest weight modules of highest weight $\lambda$ are both isomorphic to $L(\lambda )$, so equal highest weights force isomorphism.

Step 1.3 proves that an isomorphism forces equal highest weights and step 3.1 proves that equal highest weights force isomorphism, so the two directions of the equivalence are established.