The following is an explicit argument building on the knowledge of the finite-dimensional irreducible representation of ${\mathfrak g}$. At its heart is the non-degeneracy of the Shapovalov-form and the description of its determinant, but I tried to keep the exposition elementary.
Setup: Let ${\mathfrak g}={\mathfrak n}^-\oplus{\mathfrak h}\oplus{\mathfrak n}^+$ be a triangular decomposition of ${\mathfrak g}$ with respect to a Cartan subalgebra ${\mathfrak h}$ of ${\mathfrak g}$ and a choice of positive roots $\Phi^+\subset{\mathfrak h}^{\ast}$. Further, let ${\mathfrak b}:={\mathfrak h}\oplus{\mathfrak n}^+$ be the associated Borel subalgebra. Finally, recall the PBW decomposition ${\mathscr U}{\mathfrak g}\cong{\mathscr U}{\mathfrak n}^-\otimes{\mathscr U}{\mathfrak h}\otimes{\mathscr U}{\mathfrak n}^+$.
It is known (and not hard to show) that every finite-dimensional irreducible representation of ${\mathfrak g}$ is uniquely of the form $L(\lambda)=M(\lambda)/N(\lambda)$, where $\lambda\in{\mathfrak h}^{\ast}$ is dominant integral, i.e. $\lambda(h_\alpha)\in{\mathbb Z}_{\geq 0}$ for all $\alpha\in\Phi^+$, and $M(\lambda) := {\mathscr U}{\mathfrak g}\otimes_{{\mathscr U}{\mathfrak b}} {\mathbb C}_\lambda$ for the $1$-dimensional ${\mathscr U}{\mathfrak b}$-module ${\mathbb C}_\lambda$ given by ${\mathfrak n}^+{\mathbb C}_\lambda = \{0\}$ and ${\mathfrak h}$ acting on ${\mathbb C}_\lambda$ via $\lambda$.
It is important to get the idea of how $M(\lambda)$ and $L(\lambda)$ come about geometrically: The weight space diagram of $M(\lambda)$ is a downwards directed cone rooted in $\lambda$, while the one of $L(\lambda)$ is its largest symmetric subset with respect to the Weyl group action. See here, for example, where the dotted lines indicate the weight cone of $M(\lambda)$, and the solid area is where the weights of $L(\lambda)$ live.
Let's consider the point separation for elements of ${\mathscr U}{\mathfrak n}^-$ first. For those, their action on $M(\lambda)$ is very simple: As a ${\mathscr U}{\mathfrak n}^-$-module, $M(\lambda)\cong {\mathscr U}{\mathfrak n}^-$ with $1\otimes 1\mapsto 1$ because ${\mathscr U}{\mathfrak g}\cong{\mathscr U}{\mathfrak n}^-\otimes{\mathscr U}{\mathfrak h}\otimes{\mathscr U}{\mathfrak n}^+$ by PBW. So no non-zero element of ${\mathscr U}{\mathfrak n}^-$ acts trivially on $M(\lambda)$, because it doesn't kill the highest weight vector $1\otimes 1$. The idea is now to make $\lambda\gg 0$ large enough, for any fixed element of ${\mathscr U}{\mathfrak n}^-\setminus\{0\}$, so that this argument can be carried over to $L(\lambda)$, showing that the element under consideration doesn't annihilate the highest weight vector. Intuitively, this should be possible, because the larger $\lambda$ gets, the further 'away from' $\lambda$ does the submodule $N(\lambda)$ start that is annihilated from $M(\lambda)$ when passing to $L(\lambda)$.
Starting to be precise, you have the following:
Proposition: For any simple root $\alpha\in\Delta$, there is a unique embedding $M(s_\alpha\cdot\lambda)\subset M(\lambda)$, and $$L(\lambda)=M(\lambda)/\sum_{\alpha\in\Delta} M(s_\alpha\cdot\lambda).$$
NB: Pursuing this further, you get the BGG resolution of $L(\lambda)$ in terms of $M(w\cdot \lambda)$, with $w\in W$ in the $l(w)$-th syzygy.
Corollary: If $\mu\preceq\lambda$ (i.e. $\lambda-\mu\in{\mathbb Z}_+\Phi^+$, so $\mu$ is in the cone below $\lambda$) but $\lambda - \mu = \sum_{\alpha\in\Delta} c_\alpha \alpha$ with $c_\alpha < \lambda(h_\alpha)$ for all $\alpha\in\Delta$, then the projection $M(\lambda)_\mu\twoheadrightarrow L(\lambda)_\mu$ is an isomorphism.
In other words, it is only in the union of the 'shifted' cones rooted at $s_\alpha\cdot\lambda$ that $L(\lambda)$ starts looking different from $M(\lambda)$. This should be somewhat intuitive.
From that we get separation of points as follows:
Corollary: Let $\theta = \sum_{\alpha\in\Delta} c_\alpha \alpha$ with $c_\alpha\in{\mathbb Z}^+$, and suppose $y\in{\mathscr U}{\mathfrak n}^-_{-\theta}$; that is, $x$ is a sum of products $y_{\alpha_{i_1}}\cdots y_{\alpha_{i_k}}$ such that $\theta = \sum_i \alpha_{i_j}$. Then for any $\lambda\in{\mathfrak h}^{\ast}$ with $\lambda(h_\alpha)\in{\mathbb Z}^{> c_\alpha}$ for all $\alpha\in\Delta$, $y.v_\lambda\neq 0$ for the highest weight vector $v_\lambda$ of $L(\lambda)$. In particular, $xy$ doesn't act trivially on $L(\lambda)$.
Proof: If $\tilde{v}_\lambda$ denotes the highest weight vector of $M(\lambda)$, then by the previous proposition we have $y.\tilde{v}_\lambda\in M(\lambda)\setminus N(\lambda)$. In particular, $x$ acts nontrivially on the image $v_\lambda$ of $\tilde{v}_\lambda$ in $L(\lambda)$.
Corollary: Let $\theta$, $y\in{\mathscr U}{\mathfrak n}^-_{-\theta}$ and $\lambda$ be as before. Then there exists some $x\in{\mathscr U}{\mathfrak n}^+_{\theta}$ such that $(xy)_0(\lambda)\neq 0$, where $(xy)_0\in {\mathscr U}{\mathfrak h}\cong {\mathscr P}({\mathfrak h}^{\ast})$ is the projection of $xy\in{\mathscr U}{\mathfrak g}_{0}$ onto ${\mathscr U}{\mathfrak h}\subset {\mathscr U}{\mathfrak g}_{0}$ with respect to the PBW decomposition.
Here, we used that ${\mathscr U}{\mathfrak h}\cong {\mathfrak S}({\mathfrak h})\cong {\mathscr P}({\mathfrak h}^{\ast})$ can be viewed as the algebra of polynomial functions on ${\mathfrak h}^{\ast}$.
Proof: Since $y.v_\lambda\neq 0$ in $L(\lambda)$ and $L(\lambda)$ is simple, we have $L(\lambda)={\mathscr U}{\mathfrak g}.y.v_\lambda={\mathscr U}{\mathfrak n}^-{\mathscr U}{\mathfrak b}.y.v_\lambda$. In particular, there exists $x\in {\mathscr U}{\mathfrak n}^+$ such that $(xy).v_\lambda\neq 0$ in $L(\lambda)_\lambda$. For such $x$, we must have $(xy)_0\neq 0$ since the $({\mathscr U}{\mathfrak g}){\mathfrak n}^+$-component of $xy$ acts trivially on the highest weight vector $v_\lambda$. Finally, note that $z\in{\mathscr U}{\mathfrak h}$ acts on $v_\lambda$ by $z(\lambda)$.
In the previous corollary, the roles of $x$ and $y$ can be reversed:
Corollary: For $\theta$, $\lambda$ as before and $x\in {\mathscr U}{\mathfrak n}^+_{\theta}$, there exists an $y\in {\mathscr U}{\mathfrak n}^-_{-\theta}$ such that $(xy)_0(\lambda)\neq 0$.
Proof: Apply the corollary to $\tau(y)\in {\mathscr U}{\mathfrak n}^-_{-\theta}$, where $\tau:{\mathscr U}{\mathfrak g}^{\text{opp}}\to{\mathscr U}{\mathfrak g}$ is the auto-involution of ${\mathscr U}{\mathfrak g}$ swapping ${\mathfrak n}^+$ and ${\mathfrak n}^-$.
Theorem (Separation of Points): For any $z\in {\mathscr U}{\mathfrak g}\setminus\{0\}$ there exists a finite-dimensional $L(\lambda)$ such that $z.L(\lambda)\neq 0$.
Proof: Assume $z=\sum_\theta y_\theta h_\theta x_\theta$ where $x_\theta\in{\mathscr U}{\mathfrak n}^+_\theta$ and $y_\theta\in{\mathscr U}{\mathfrak n}^-$, $h_\theta\in{\mathscr U}{\mathfrak h}$; in other words, you group PBW terms by the weight on the ${\mathfrak n}$-side. Now, consider $\theta$ maximal w.r.t. the ordering $\lambda\preceq\mu:\Leftrightarrow \mu-\lambda\in{\mathbb Z}^+\Phi^+$ such that $y_\theta h_\theta x_\theta$ nonzero. Then, we know from our previous work that there's some $\lambda\gg 0$ such that for any $\lambda^{\prime}$ such $\lambda\preceq\lambda^{\prime}$ there exists some $y^{\prime}_\theta\in{\mathscr U}{\mathfrak n}^-_{-\theta}$ (depending on $\lambda^{\prime}$) such that $(x_\theta y^{\prime}_\theta)_0(\lambda^{\prime})\neq 0$. Picking $\lambda^{\prime}$ large enough, we may assume that also $h_\theta(\lambda^{\prime})\neq 0$; this is because the polynomial $h_\theta\in {\mathscr P}{\mathfrak h}\cong{\mathscr P}({\mathfrak h}^{\ast})$ cannot vanish on the shifted half-lattice $\lambda + {\mathbb Z}^+\Phi^+$. Putting everything together, in $L(\lambda^{\prime})$ we then have $(y_\theta h_\theta x_\theta).(y^{\prime}_\theta v_{\lambda^{\prime}}) = h_\theta(\lambda^{\prime}) (x_\theta y^{\prime}_\theta)_0(\lambda^{\prime}) y_\theta v_{\lambda^{\prime}}\neq 0$, where for the last step we potentially have to enlarge $\lambda^{\prime}$ again. What about the other summands in $z$? They all annihilate $y^{\prime}_\theta v_{\lambda^{\prime}}$ because of the maximality of $\theta$.
To sum up the discussion in the comment into an answer: It is indeed true that for each $A \in \mathfrak g$, $\rho_W(A)$ is a homomorphism of $S_N$-modules. However, Schur's Lemma talks about homomorphisms of simple (a.k.a irreducible) $S_N$-modules, and as OP said, generally $W$ is not irreducible, but decomposes as a direct sum of irreducible $X_i$.
It is still possible to infer something from Schur's Lemma, namely:
For each such irreducible $X_i$, the restriction of $\rho_W(A)$ to $X_i$ is either
- $\rho_W(A)_{\vert X_i} = 0$
or
- $\rho_W(A)_{\vert X_i}$ induces an isomorphism (of $S_N$-modules) to another $X_j$ (where in general $j\neq i$).
The part of Schur's lemma whose sloppy interpretation is that "our endomorphism is a scalar" actually says that the endomorphism ring of a simple $S_N$-module is a skew field, finite dimensional over the ground field $k$ we've tacitly been working with all the time. If that field was $\mathbb C$, then that endomorphims ring is necessarily $\mathbb C$, but if our field was $\mathbb R$, it could in principle be $\mathbb R, \mathbb C$ or $\mathbb H$. However, I think that for the symmetric group $S_N$ it's actually known (cf. MO/10635) that all real (or even rational) representations have Schur index $1$ ($\Leftrightarrow$ Frobenius-Schur indicator $1$), meaning that indeed $End_{\mathbb R[S_N]}(X_i) \simeq \mathbb R$ and hence
- if in the second case above $i=j$, then $\rho_W(A)_{\vert X_i}$ is given by multiplication with some $\lambda_i \in \mathbb R^*$.
To see this in an example, let $k=\mathbb R, \mathfrak g = \mathfrak{sl}_2(\mathbb R), V=$ the standard representation of $\mathfrak g$ on $\mathbb R^2$, and $N=2$. Then $W= V^{\otimes 2}$ as $\mathbb R[S_2]$-module decomposes into four $1$-dimensional components; namely, set $x_i = e_i \otimes e_i $ for $i=1,2$, $x_3 = e_1 \otimes e_2 +e_2\otimes e_1$, and $x_4 = e_1 \otimes e_2 -e_2\otimes e_1 $, and let $X_i := \mathbb Rx_i$. Then $X_4$ is the alternating (sign) representation, the three other $X_i$ are isomorphic to the trivial representation.
Now e.g. for $A=\pmatrix{0 &1\\0&0}$ we have $\rho_W(A)_{\vert X_i} = \begin{cases} 0 \text{ if } i=1,4 \\ x_2 \mapsto x_3 \text{ if } i=2, \text{ giving an iso } X_2 \simeq X_3 \\ x_3 \mapsto 2 x_1 \text{ if } i=3, \text{ giving an iso } X_3 \simeq X_1 \end {cases}$
whereas for $H=\pmatrix{1 &0\\0&-1}$ we have $\rho_W(H)_{\vert X_i} = \begin{cases} x_1 \mapsto 2x_1 \text{ if } i=1, \text{ i.e. } \lambda_1=2 \\x_2 \mapsto -2x_2 \text{ if } i=2, \text{ i.e. } \lambda_2=-2 \\ 0 \text{ if } i=3,4\end {cases}$
etc.
Best Answer
The comultiplication $\Delta:U(\mathfrak{g})\to U(\mathfrak{g})\otimes U(\mathfrak{g})$ is given by $x\mapsto x\otimes 1+1\otimes x$ only for $x\in\mathfrak{g}$, not for arbitrary $x\in U(\mathfrak{g})$. This map is then extended to all of $U(\mathfrak{g})$ to be an algebra homomorphism (using the universal property of $U(\mathfrak{g})$). So this means that for $x,y\in\mathfrak{g}$, $\Delta(xy)$ is actually defined as $\Delta(xy)=\Delta(x)\Delta(y)=(x\otimes 1+1\otimes x)(y\otimes 1+1\otimes y)$, not as $xy\otimes 1+1\otimes xy$.