I'm reading the proof of lemma 19.1 in Carter's Lie algebras of finite and affine type, which states that the category $\mathcal{O}$ is closed under tensor products. I'm having a bit of trouble understanding a couple of steps in the proof.
We first suppose $V_1,V_2\in\mathcal{O}$ (the category $\mathcal{O}$). Then $V_1=\bigoplus_{\lambda\in H^*}(V_1)_\lambda$ and $V_2=\bigoplus_{\lambda\in H^*}(V_2)_\lambda$ with $(V_1)_\lambda, (V_2)_\lambda$ finite-dimensional and it's weights bounded by a finite subset of $H^*$. Thus we have $V_1\otimes V_2=\bigoplus_{\lambda_1,\lambda_2\in H^*}((V_1)_{\lambda_1}\otimes(V_2)_{\lambda_2})$. Now one have $((V_1)_{\lambda_1}\otimes(V_2)_{\lambda_2})\subset (V_1\otimes V_2)_{\lambda_1+\lambda_2}$ which is also clear to me, but why does it imply that $V_1\otimes V_2=\bigoplus_{\lambda\in H^*}(V_1\otimes V_2)_\lambda$ with $(V_1\otimes V_2)_\lambda=\sum_{\lambda_1+\lambda_2=\lambda}((V_1)_{\lambda_1}\otimes(V_2)_{\lambda_2})$. How does the inclusion implies that $(V_1\otimes V_2)_\lambda=\sum_{\lambda_1+\lambda_2=\lambda}((V_1)_{\lambda_1}\otimes(V_2)_{\lambda_2})$? Wouldn't one need to show that $(V_1\otimes V_2)_\lambda\subset\sum_{\lambda_1+\lambda_2=\lambda}((V_1)_{\lambda_1}\otimes(V_2)_{\lambda_2})$?
Now if there seems to be an easier proof of why the category $\mathcal{O}$ is closed under the tensor product, I'm very happy to see it (or maybe a reference to it).
It's worthwhile to work slightly more generally, since this comes up whenever you have modules graded by an abelian group.
First, we have that tensor products commute with direct sums, so $$\left(\bigoplus_{i\in I} M_i \right) \otimes \left( \bigoplus_{j\in J} N_j \right)\cong \bigoplus_{(i,j)\in I\times J} M_i \otimes N_j.$$ For a quick proof, tensor products are left adjoint to hom, so they commute with arbitary colimits, and in particular coproducts (which is what direct sums are).
If the indexing sets are both some abelian group $A$, then we can further decompose $$\left(\bigoplus_{a\in A} M_a \right) \otimes \left( \bigoplus_{b\in A} N_b \right)\cong \bigoplus_{(a,b)\in A\times A} M_i \otimes N_j\cong \bigoplus_{c\in A}\bigoplus_{a+b=c} M_a\otimes N_b.$$
At a more concrete level, given an element of the left hand side, we can write it as a sum of pure tensor products (things of the form $m\otimes n$), and we can decompose each $m$ and $n$ into homogenous components and then split things apart into a sum of pure tensors of homogenous elements.
Once we have this abstract decomposition, let's show that it's actually the decomposition of $V_1\otimes V_2$ into weight spaces. It's a straight forward calculation that $(V_1)_{\lambda_1}\otimes (V_2)_{\lambda_2}$ is homogenous of weight $\lambda_1+\lambda_2$, and so we have an inclusion $$\bigoplus_{\lambda_1+\lambda_2=\lambda} (V_1)_{\lambda_1}\otimes (V_2)_{\lambda_2}\subset (V_1\otimes V_2)_{\lambda}$$, but summing over all $\lambda$ and using the above isomorphism, we have $$V_1 \otimes V_2 \cong \bigoplus_{\lambda}\bigoplus_{\lambda_1+\lambda_2=\lambda} (V_1)_{\lambda_1}\otimes (V_2)_{\lambda_2}\subset \bigoplus_{\lambda}(V_1\otimes V_2)_{\lambda}\cong V_1\otimes V_2.$$
Moreover, since the isomorphisms come from the natural inclusions, this forces the inclusion in the middle to be equality.
If you wish to abstract out the idea here, if you have $A_i\subset B_i$ for all $i\in I$, and if $\bigoplus_i A_i = \bigoplus B_i$, then $A_i=B_i$ for all $i$.
Again, the reason this abstraction is what is happening is because the isomorphisms above are coming from the natural inclusions.
As far as addressing local finiteness, given a sum $v=\sum v_{\alpha}$, we have an inclusion $\mathfrak nv \subset \sum \mathfrak n v_{\alpha}$, and the above isomorphism tells us we can write any $v\in V_1\otimes V_2$ as a sum of elements of the form $v_{\lambda_1}\otimes v_{\lambda_2}$ where $v_1$ and $v_2$ are homogenous of weight $\lambda_1$ and $\lambda_2$ respectively, but $\mathfrak n(v_1\otimes v_2)\subset (\mathfrak n v_1)\otimes (\mathfrak n v_2)$, which is a tensor product of finite dimensional vector spaces, and therefore finite dimensional. Since a finite sum of finite dimensional vector spaces is finite dimensional, the result follows.