If $\mathfrak{g}=\bigoplus\mathfrak{g}_i$ is a semisimple Lie algebra, why does $\mathfrak{h}=\bigoplus\left(\mathfrak{h}\cap\mathfrak{g}_i\right)$? There is this property about Cartan subalgebras that is not clear to me. 
Suppose $\mathfrak{g}$ is a semisimple Lie algebra. Then I know we can decompose it uniquely as $\mathfrak{g}=\bigoplus\mathfrak{g}_i$, where the $\mathfrak{g}_i$ are simple ideals. If $\mathfrak{h}$ is a Cartan subalgebra (maximal toral subalgebra), then $\mathfrak{h}=\bigoplus(\mathfrak{h}\cap\mathfrak{g}_i)$. The $\supseteq$ direction is clear. 
Why does the other follow? I thought of something like this, take $x\in\mathfrak{h}$, and write $x=\sum x_i$ for $x_i\in\mathfrak{g}_i$. I want to show $x_i\in\mathfrak{h}$ for each $i$. I did this by induction on the number of nonzero summands. If there is only one summand so $x=x_i$, the claim is clear. If there's more than one summand, I'm not sure how to reduce it to imply the induction, if this is the right idea.
 A: Suppose that $K$ is the base field.  Fix $h\in\mathfrak{h}$.  Let $h_i\in\mathfrak{g}_i$ be such that $h=\sum\limits_i\,h_i$ (the $h_i$'s are unique).  We claim that $h_i\in\mathfrak{h}$ for all $i$.  Let $t\in\mathfrak{h}$.  Write $t=\sum\limits_{i}\,t_i$ with $t_i\in\mathfrak{g}_i$.  For $i\neq j$, $$\left[t_i,h_j\right]\in\mathfrak{g}_i\cap\mathfrak{g}_j=\{0\}\,,$$ so $\left[t_i,h_j\right]=0$.  Ergo, $$\sum\limits_{i}\,\left[t_i,h_i\right]=[t,h]=0\,.$$  Since $\left[t_i,h_i\right]\in\mathfrak{g}_i$, we conclude that $\left[t_i,h_i\right]=0$ for every $i$.  Thus, $\left[h_i,t\right]=\left[h_i,t_i\right]=0$ for every $t\in\mathfrak{h}$.  
Because $h$ is a semisimple element of $\mathfrak{g}$, $h_i$ is also a semisimple element of $\mathfrak{g}$ (to be proven below).  If $h_i\notin\mathfrak{h}$, then $\mathfrak{h}\oplus K\mathfrak{h}_i$ is a strictly larger toral subalgebra containing $h$, which is a contradiction.  Thus, $h_i\in \mathfrak{h}$ for all $i$, so $h_i\in\mathfrak{h}\cap\mathfrak{g}_i$.  Therefore, $$h=\sum\limits_i\,h_i\in\bigoplus\limits_i\,\left(\mathfrak{h}\cap\mathfrak{g}_i\right)\,,$$ whence $\mathfrak{h}\subseteq \bigoplus\limits_i\,\left(\mathfrak{h}\cap\mathfrak{g}_i\right)$.  Since it is trivial that $\mathfrak{h}\supseteq \bigoplus\limits_i\,\left(\mathfrak{h}\cap\mathfrak{g}_i\right)$, $\mathfrak{h}= \bigoplus\limits_i\,\left(\mathfrak{h}\cap\mathfrak{g}_i\right)$, as required.
To show that $h_i$ is a semisimple element of $\mathfrak{g}$, we let $V^\lambda$ be the eigenspace of $\text{ad}_\mathfrak{g}(h)$ with the eigenvalue $\lambda\in K$.  For each $i$, take $V_i^\lambda$ to be the intersection $V^\lambda\cap\mathfrak{g}_i$.  Trivially, $\bigoplus\limits_{\lambda\in K}\,V^\lambda_i=\mathfrak{g}_i$.  However, as $V^\lambda_i$ is the eigenspace with eigenvalue $\lambda$ of $\text{ad}_{\mathfrak{g}_i}\left(h_i\right)$, we see that $h_i$ is a semisimple element of $\mathfrak{g}_i$.  Now, define $\tilde{V}^\lambda_i:=V^\lambda_i$ if $\lambda\neq 0$, and $\tilde{V}^0_i:=V^0_i\oplus\left(\bigoplus\limits_{j\neq i}\,\mathfrak{g}_j\right)$.  Then, $\tilde{V}^\lambda_i$ is the eigenspace of $\text{ad}_\mathfrak{g}\left(h_i\right)$ with eigenvalue $\lambda$.  Since $\bigoplus\limits_{\lambda \in K}\,\tilde{V}^\lambda_i=\mathfrak{g}$, we conclude that $h_i$ is a semisimple element of $\mathfrak{g}$.
A: This is true in any finite-dimensional Lie algebra over any field.
Recall that a Cartan subalgebra is a nilpotent subalgebra equal to its normalizer. In particular it's maximal nilpotent.
If we have $\mathfrak{g}=\prod_{i=1}^n\mathfrak{g}_i$ a direct product of such Lie algebras and $\mathfrak{h}$ is nilpotent and $\mathfrak{h}_i$ is its projection to $\mathfrak{g}_i$, then $\prod_{i=1}^n\mathfrak{h}_i$ is nilpotent and contains $\mathfrak{h}$. In particular, if $\mathfrak{h}$ is maximal nilpotent, then we deduce that $\mathfrak{h}=\prod_{i=1}^n\mathfrak{h}_i$, that is, $\mathfrak{h}$ is stable under taking projection to the direct summands. This applies in particular when $\mathfrak{h}$ is Cartan.
