It is true.

1st case: $G$ is linear, i.e., isomorphic to a closed subgroup of matrices. Here I don't need simple connectedness — I even need not to assume it, and I only assume that $G_1,G_2$ are immersed connected subgroups, possibly not closed.

Note that the case when $G_1,G_2$ are Zariski-closed is standard, as has already be mentioned. But I'm not assuming this.

First a simple group-theoretic fact: $G_1, G_2$ both normalize $C=[G_1,G_2]$. Indeed, write the commutator as $[g,h]=ghg^{-1}h^{-1}$ (the choice of convention does not affect the definition of $[G_1,G_2]$). Then $[g,st]=[g,s]s[g,t]s^{-1}$.

Next, it is a known fact that the commutator subgroup of every Lie subalgebra is algebraic. Hence, after modding out, we can suppose that $C$ is abelian.

At this point, we can enlarge $G$ and suppose $G=\mathrm{GL}_n(\mathbf{C})$.

Hence, writing $C=[G_1,G_2]$, taking $g\in G_1$, $s,t\in G_2$, we see that $s[g,t]s^{-1}\in C$. Viewing this for each $s$, and all $g,t$, we see that $s$ normalizes $C$. Since this is true for each $s$, we see that $G_2$ normalizes $C$. Similarly $G_1$ normalizes (e.g., because $[G_1,G_2]=[G_2,G_1]$).

Let $U$ be the (complex) Zariski-closure of $C$; this is a Zariski-closed connected abelian group. decompose the $C$-action into weight blocks $V_\chi$ ($\chi\in\mathrm{Hom}(U,\mathbf{C}^*)$). In the block $V_\chi$, $U$ acts by the scalar $V_\chi$ times some unipotent element (said otherwise, in $V_\chi$, we can triangulate the $U$-action so that $u$ acts as an upper triangular matrix with constant diagonal $\chi(u)$).

Since $G_i$ normalizes $C$, it normalizes $U$, and hence preserves the weight decomposition, i.e., preserves each $V_\chi$. Hence $[G_1,G_2]$ has determinant one in each $V_\chi$. But the determinant of $u\in U$ on $V_\chi$ is $\chi(u)^{\dim(V_\chi)}$. Since $C$ is Zariski-dense in $U$, this has to be $=1$. Since $U$ is connected, this is constant only if $\chi=1$. Thus $\mathbf{C}^n=V_1$. In other words, $U$ is upper unipotent. Then every real Lie subalgebra of $\mathfrak{u}$ defines a (real Zariski-closed) subgroup. Hence $[G_1,G_2]=C$ is real Zariski closed (and unipotent).

In general (i.e. going back before the reduction to $C$ abelian), we reach the conclusion that $[G_1,G_2]$ is real Zariski-closed and has a unipotent abelianization (i.e., its reductive Levi factor is semisimple).

General case: every simply connected Lie group $G$ has a quotient $G/Z$ by a discrete central subgroup that is linear. Then $[G_1,G_2]$ is a lift of $[G_1Z/Z,G_2Z/Z]$, which is closed (using the 1st case, where I need not to assume that the subgroups are closed!), and hence is closed as well.

It is true.

1st case: $G$ is linear, i.e., isomorphic to a closed subgroup of matrices. Here I don't need simple connectedness — I even need not to assume it, and I only assume that $G_1,G_2$ are immersed connected subgroups, possibly not closed.

Note that the case when $G_1,G_2$ are Zariski-closed is standard, as has already be mentioned. But I'm not assuming this.

First a simple group-theoretic fact: $G_1, G_2$ both normalize $C=[G_1,G_2]$. Indeed, write the commutator as $[g,h]=ghg^{-1}h^{-1}$ (the choice of convention does not affect the definition of $[G_1,G_2]$). Then $[g,st]=[g,s]s[g,t]s^{-1}$. Hence, writing $C=[G_1,G_2]$, taking $g\in G_1$, $s,t\in G_2$, we see that $s[g,t]s^{-1}\in C$. Viewing this for each $s$, and all $g,t$, we see that $s$ normalizes $C$. Since this is true for each $s$, we see that $G_2$ normalizes $C$. Similarly $G_1$ normalizes (e.g., because $[G_1,G_2]=[G_2,G_1]$).

Next, it is a known fact that the commutator subgroup of every Lie subalgebra is algebraic. Hence, after modding out, we can suppose that $C$ is abelian.

At this point, we can enlarge $G$ and suppose $G=\mathrm{GL}_n(\mathbf{C})$.

Let $U$ be the (complex) Zariski-closure of $C$; this is a Zariski-closed connected abelian group. decompose the $C$-action into weight blocks $V_\chi$ ($\chi\in\mathrm{Hom}(U,\mathbf{C}^*)$). In the block $V_\chi$, $U$ acts by the scalar $V_\chi$ times some unipotent element (said otherwise, in $V_\chi$, we can triangulate the $U$-action so that $u$ acts as an upper triangular matrix with constant diagonal $\chi(u)$).

Since $G_i$ normalizes $C$, it normalizes $U$, and hence preserves the weight decomposition, i.e., preserves each $V_\chi$. Hence $[G_1,G_2]$ has determinant one in each $V_\chi$. But the determinant of $u\in U$ on $V_\chi$ is $\chi(u)^{\dim(V_\chi)}$. Since $C$ is Zariski-dense in $U$, this has to be $=1$. Since $U$ is connected, this is constant only if $\chi=1$. Thus $\mathbf{C}^n=V_1$. In other words, $U$ is upper unipotent. Then every real Lie subalgebra of $\mathfrak{u}$ defines a (real Zariski-closed) subgroup. Hence $[G_1,G_2]=C$ is real Zariski closed (and unipotent).

In general (i.e. going back before the reduction to $C$ abelian), we reach the conclusion that $[G_1,G_2]$ is real Zariski-closed and has a unipotent abelianization (i.e., its reductive Levi factor is semisimple).

General case: every simply connected Lie group $G$ has a quotient $G/Z$ by a discrete central subgroup that is linear. Then $[G_1,G_2]$ is a lift of $[G_1Z/Z,G_2Z/Z]$, which is closed (using the 1st case, where I need not to assume that the subgroups are closed!), and hence is closed as well.