$\renewcommand{\g}{{\mathfrak g}} $ We say that a Lie algebra $\g$ is perfect if $[\g,\g]=\g$.
Question. Does there exist a finite dimensional, perfect Lie algebra $\g$ over $\Bbb C$ with nontrivial center?
If the answer is "Yes", I would like to see a nice example.
On
$\def\SL{{\rm SL}} \def\GL{{\rm GL}} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Der}{Der} \DeclareMathOperator{\Lie}{Lie} \def\e{{\mathfrak e}} \def\g{{\mathfrak g}} \def\h{{\mathfrak h}} \def\s{{\mathfrak s}} \def\z{{\mathfrak z}} \def\sl{{\mathfrak{sl}}} \def\gl{{\mathfrak{gl}}} $ The answer is Yes; see the example below. See also the papers by Rutwig Campoamor-Stursberg: The structure of the invariants of perfect Lie algebras. J. Phys. A 36 (2003), no. 24, 6709–6723 and The structure of the invariants of perfect Lie algebras. II. J. Phys. A 37 (2004), no. 11, 3627–3643 (thanks to @Callum for the reference).
Derivations. Let $k$ be a field, and $\g$ be a finite dimensional Lie algebra over $k$. A derivation of $\g$ is a linear map $\delta\colon \g\to \g$ such that $$\delta\big([x,y]\big)=[\delta(x),y] + [x,\delta(y)]\quad\ \text{for all}\ \, x,y \in\g.$$ The derivations of $\g$ form a vector space $\Der\g$, which is a Lie subalgebra of $\gl(\g)$. Consider the automorphism group $\Aut\g$, which is a linear algebraic group, an algebraic subgroup of $\GL(\g)$. Then $\Der\g=\Lie(\Aut\g)$; see this answer of Qiaochu Yuan.
Semi-direct sum of Lie algebras. Let $\g$ and $\h$ be Lie algebras, and let $\phi\colon \g\to\Der\h$ be a homomorphism. Following Wikipedia: Lie algebra extension, we construct a semi-direct sum $\h\oplus_\phi \g$. We construct a skew-symmetric bracket on the vector space $\h\oplus\g$: $$ \big[(h_1,g_1),\,(h_2,g_2)\big]=\big([h_1,h_2]+\phi_{g_1}(h_2)-\phi_{g_2}(h_1),\, [g_1,g_2]\big)$$ for $h_1,h_2\in \h,\ g_1,g_2\in \g$. We check the Jacobi identity. We calculate: \begin{align*} \big[\, [&(h_1,g_1),(h_2,g_2)],\, (h_3,g_3)\,\big] =\big([[h_1,h_2],h_3]+[\phi_{g_1}(h_2),h_3]-[\phi_{g_2}(h_1),h_3]\\ & +\phi_{[g_1,g_2]}(h_3)-\phi_{g_3}([h_1,h_2])-\phi_{g_3}(\phi_{g_1}(h_2)) +\phi_{g_3}(\phi_{g_2}(h_1)),\, [[g_1,g_2],g_3]\big)\\ &=\big(\,[[h_1,h_2],h_3]+[\phi_{g_1}(h_2),h_3]-[\phi_{g_2}(h_1),h_3]+\phi_{g_1}(\phi_{g_2}(h_3))-\phi_{g_2}(\phi_{g_1}(h_3))\\ & -[\phi_{g_3}(h_1),h_2]+[\phi_{g_3}(h_2),h_1]-\phi_{g_3}(\phi_{g_1}(h_2)) +\phi_{g_3}(\phi_{g_2}(h_1)),\, [[g_1,g_2],g_3]\,\big). \end{align*} Using the Jacobi identities for $\g$ and $\h$: \begin{gather*} [[h_1,h_2],h_3]+[[h_2,h_3],h_1] +[[h_3,h_1],h_2]=0,\\ [[g_1,g_2],g_3]+[[g_2,g_3],g_1] +[[g_3,g_1],g_2]=0, \end{gather*} we obtain \begin{multline*} \big[\, [(h_1,g_1),(h_2,g_2)],\, (h_3,g_3)\,\big]+ \big[\, [(h_2,g_2),(h_3,g_3)],\, (h_1,g_1)\,\big]\\ +\big[\, [(h_3,g_3),(h_1,g_1)],\, (h_2,g_2)\,\big]=0 \end{multline*} (all the terms cancel out), which gives the Jacobi identity for $\h\oplus_\phi\g$.
Example 1. Let $k$ be a field, $V=k^2$, $W=k$, $\psi\colon V\to W$ the standard non-degenerate skew-symmetric bilinear form.
We set $\h=V\oplus W$ and define a skew-symmetric bracket $$\h\times \h\to W\hookrightarrow \h,\quad\, \big[(v_1,w_1), (v_2,w_2)\big]=\big(0, \psi(v_1,v_2)\big).$$ We have $[\h,\h]=W$, whence $[[\h,\h],\h]=0$, and the Jacobi identity for $\h$ is trivially satisfied. Thus $\h$ is a (two-step nilpotent) Lie algebra with center $W$.
Set $S=\SL_2$ and $\s={\rm Lie}(S)=\sl_2$. The algebraic group $S$ acts on the vector space $V=k^2$, and it acts on the vector space $\h$ by $$ s, (v,w)\mapsto (s(v), w)\quad\ \text{for}\ \,s\in S,\ v\in V,\ w\in W.$$ The action on $\h$ preserves the Lie bracket in $\h$, and so we obtain a homomorphism $f\colon S\to \Aut \h$. Differentiating, we obtain a homomorphism $$\phi:=df\colon \s\to \Lie(\Aut\h)=\Der \h.$$ An easy calculation shows that $$\phi_s(v,w)=(s(v),0)\quad\ \text{for}\ \, s\in \s,\ v\in V,\ w\in W.$$ As above, we define a Lie algebra $$\g=\h\oplus_\phi\s$$ with the bracket $$ \big[(v_1,w_1,s_1),(v_2,w_2,s_2)\big]=\big(s_1(v_2)-s_2(v_1),\, \psi(v_1,v_2),\, [s_1,s_2]\big).$$
Proposition. The constructed $6$-dimensional Lie algebra $\g$ is perfect with one-dimensional center $\z(\g)=W$.
Proof. From the formula $$\big[(0,0,s_1),(0,0,s_2)\big]=\big(0,0,[s_1,s_2]\big)$$ we see that $[\g,\g]\supseteq \s$. From the formula $$\big[(0,0,s_1),(v_2,0,0)\big]=\big( s_1(v_2),0,0\big)$$ we see that $[\g,\g]\supseteq V$. From the formula $$\big[(v_1,0,0),(v_2,0,0)\big]=\big(0,\psi(v_1,v_2),0\big)$$ we see that $[\g,\g]\supseteq W$. Thus $[\g,\g]=\g$, that is, $\g$ is perfect.
From the formula $$ \big[(v_1,w_1,s_1),(0,w_2,0)\big]=(0,\, 0,\, 0)$$ we see that $W\subseteq\z(\g)$.
From the formula $$\big[(0,0,s_1),(v_2,w_2,s_2)\big]=\big( s_1(v_2),0,[s_1,s_2]\big)$$ we see that the centralizer of $\s$ in $\g$ is $W$. Thus $\z(\g)\subseteq W$ and $\z(\g)=W$.
We conclude that $\g$ is perfect and has one-dimensional center $W$.
EDIT: A more genral example, containing the example of @Callum.
Example 2. Let $\s$ be a semisimple Lie algebra, $V$ be a finite dimensional representation of $\s$ (maybe reducible) admitting a nontrivial invariant skew-symmetric bilinear form $\psi\colon V\times V\to k$ and not containing the trivial representation in $k$ as a direct summand. We set $W=k$ and define a structure of Lie algebra on $\h:=V\oplus W$ by $$\big[(v_1,w_1),\,(v_2,w_2)\big]=\big(0,\psi(v_1,v_2)\big).$$ We define a homomorphism $$\phi\colon\s\to\Der\h,\quad\ \phi_s(v,w)=(s(v),0) \quad\ \text{for}\ \, s\in\s,\ v\in V,\ w\in W.$$ We check that $\phi_s$ is a derivation of $\h$. The assumption that $\psi$ is $\s$-invariant means that $$\psi(s(v_1), v_2)+\psi(v_1,s(v_2))=0\quad\ \text{for all}\ \, s\in\s,\ v_1,v_2\in V.$$ We calculate: $$\phi_s\big[(v_1,w_1),\,(v_2,w_2)\big]=\phi_s\big(0,\psi(v_1,v_2)\big)=0.$$ On the other hand, \begin{align*} \big[\phi_s(v_1,w_1), (v_2,w_2)\big]+&\big[(v_1,w_1), \phi_s(v_2,w_2)\big]\\ =&\big[(s(v_1),0), (v_2,w_2)\big]+\big[(v_1,w_1), (s(v_2),0)\big]\\ =&\big(0,\,\psi(s_1(v_1),v_2)+\psi(v_1,s(v_2))\big)=0. \end{align*} Thus $$\phi_s\big[(v_1,w_1),\,(v_2,w_2)\big] =\big[\phi_s(v_1,w_1), (v_2,w_2)\big]+\big[(v_1,w_1), \phi_s(v_2,w_2)\big],$$ as required.
We check that the linear map $$ \phi\colon \s\to \Der\h,\quad\ s\mapsto\phi_s$$ is a homomorphism of Lie algebras. Indeed, $$\phi_{[s_1,s_2]}(v,w)=\big([s_1,s_2](v),0\big)=\big(s_1(s_2(v))-s_2(s_1(v)),\,0)\big).$$ On the other hand, \begin{align*} \phi_{s_1}(\phi_{s_2}(v,w))-\phi_{s_2}(\phi_{s_1}(v,w))=&\phi_{s_1}(s_2(v),0)-\phi_{s_2}(s_1(v),0)\\ =&\big(s_1(s_2(v)),\,0\big)-\big(s_2(s_1(v)),\,0\big). \end{align*} Thus $$\phi_{[s_1,s_2]}(v,w)=\phi_{s_1}(\phi_{s_2}(v,w))-\phi_{s_2}(\phi_{s_1}(v,w)),$$ as required.
We set $\g=\h\oplus_\phi \s$. Arguing as in Example 1 and using the assumptions
that $\s$ is semisimple (hence perfect),
that the representation of $\s$ in $V$ does not contain the trivial representation
as a direct summand (whence $\big\langle s(v)\ |\ s\in S,\, v\in V\big\rangle=V\,$),
and that the skew-symmetric form $\psi$ is nontrivial
(whence $\big\{\psi(v_1,v_2)\ |\ v_1,v_2\in V\big\}=k\,$),
we conclude that $\g$ is perfect and has one-dimensional center $W$.
On
Here's a nice simple example:
Let $\frak{g}$ be a Lie algebra with Levi decomposition $\mathfrak{g} = \mathfrak{sl}_2 \oplus \frak{n}$ where $\mathfrak{n} = V \oplus V^* \oplus Z$. Here $V$ is a non-trivial irreducible $\mathfrak{sl}_2$ representation and $Z =\langle z \rangle$ is a $1$-dimensional trivial representation.
Define a bracket on $\mathfrak{sl}_2 \times \frak{n}$ by these representations and one on $\mathfrak{n} \times \frak{n}$ such that $[v, f]:= v(f)z$ and all other brackets are $0$. Note this makes $\mathfrak{n}$ a Heisenberg Lie algebra.
Then $\frak{g}$ is perfect since $[\mathfrak{sl}_2,\mathfrak{sl}_2] = \mathfrak{sl}_2$, $[\mathfrak{sl}_2,\mathfrak{n}] = V \oplus V^*$, $[\mathfrak{n},\mathfrak{n}] = Z$. Then $Z$ is the centre of $\frak{g}$.
Edit: Note this example is actually distinct from the others given here despite also being the semidirect product of $\mathfrak{sl}_2$ and a Heisenberg Lie algebra. This construction only gives Heisenberg Lie algebras $\mathfrak{n}_{2n+1}$ for $n>1$ (i.e. $V$ is non-trivial). If $V$ is trivial this produces a non-perfect Lie algebra: $[\mathfrak{g},\mathfrak{g}] = \mathfrak{sl}_2 \oplus Z$.
The $6$-dimensional Lie algebra $\mathfrak{sl}_2(\Bbb C)\ltimes_{\phi} \mathfrak{n}_3(\Bbb C)$, which appears in the classification of all complex $6$-dimensional Lie algebras here is perfect and has $1$-dimensional center. Here $\mathfrak{n}_3(\Bbb C)$ is the $3$-dimensional Heisenberg Lie algebra. It seems that this is also the example from the other answer. In fact, the other Lie algebras from the classification list in dimension $6$ are either not perfect, or have trivial center.