[proofplan]
We first show that connected components of the underlying manifold of $G$ are closed, using local connectedness of manifolds. Then we use continuity of the Lie group operations to show that multiplication and inversion preserve the identity component. Finally, conjugation by an arbitrary element of $G$ is a homeomorphism sending $e$ to $e$, so it sends $G_0$ into itself; applying the same argument to the inverse conjugation gives equality and hence normality.
[/proofplan]
[step:Show that the identity component is closed in the manifold topology]
Let $C$ be any connected component of the underlying [topological space](/page/Topological%20Space) of $G$. We first verify that $C$ is open. If $x\in C$, choose a smooth chart $(U,\varphi)$ of the manifold $G$ with $x\in U$ and
\begin{align*}
\varphi:U\to \varphi(U)\subset \mathbb R^n
\end{align*}
a homeomorphism onto an open subset of $\mathbb R^n$. Since $\varphi(U)$ is open, there exists $r>0$ such that the Euclidean open ball $B(\varphi(x),r)$ is contained in $\varphi(U)$. Define
\begin{align*}
V:=\varphi^{-1}(B(\varphi(x),r)).
\end{align*}
The set $B(\varphi(x),r)$ is connected, and $\varphi^{-1}$ is continuous on $\varphi(U)$, so $V$ is connected. Since $x\in V$ and $C$ is the connected component containing $x$, maximality of connected components gives $V\subset C$. Thus every point of $C$ has an open neighbourhood contained in $C$, so $C$ is open.
Every connected component of $G$ is therefore open. Hence
\begin{align*}
G\setminus C
\end{align*}
is a union of connected components different from $C$, and is open. Therefore $C$ is closed. Applying this to the connected component $G_0$ containing $e$, we conclude that $G_0$ is closed in $G$.
[/step]
[step:Prove that the identity component is closed under multiplication]
Let
\begin{align*}
m:G\times G\to G,\qquad (a,b)\mapsto ab
\end{align*}
be the multiplication map of the Lie group $G$. Since $G$ is a Lie group, $m$ is continuous. The product space $G_0\times G_0$ is connected because it is a product of connected spaces. Therefore
\begin{align*}
m(G_0\times G_0)=G_0G_0
\end{align*}
is connected as the continuous image of a [connected space](/page/Connected%20Space).
Moreover, since $e\in G_0$, the set $G_0G_0$ contains $ee=e$. By the defining maximality of the connected component $G_0$ containing $e$, every connected subset of $G$ containing $e$ is contained in $G_0$. Hence
\begin{align*}
G_0G_0\subset G_0.
\end{align*}
Thus $G_0$ is closed under multiplication.
[guided]
We need to prove that if $a,b\in G_0$, then $ab\in G_0$. The reason connectedness is useful is that the identity component $G_0$ is the largest connected subset of $G$ containing $e$.
Define the Lie group multiplication map by
\begin{align*}
m:G\times G\to G,\qquad (a,b)\mapsto ab.
\end{align*}
By the definition of a Lie group, this map is smooth, hence continuous. The space $G_0$ is connected by definition, so the product $G_0\times G_0$ is connected. Applying continuity of $m$, the image
\begin{align*}
m(G_0\times G_0)=\{ab:a\in G_0,\ b\in G_0\}
\end{align*}
is connected.
This image also contains the identity element, because $e\in G_0$ and
\begin{align*}
m(e,e)=ee=e.
\end{align*}
Now use the defining property of the connected component $G_0$: it contains every connected subset of $G$ that contains $e$. Since $m(G_0\times G_0)$ is connected and contains $e$, we get
\begin{align*}
m(G_0\times G_0)\subset G_0.
\end{align*}
Equivalently, $ab\in G_0$ for all $a,b\in G_0$. Therefore $G_0$ is closed under multiplication.
[/guided]
[/step]
[step:Prove that the identity component is closed under inversion]
Let
\begin{align*}
i:G\to G,\qquad a\mapsto a^{-1}
\end{align*}
be the inversion map. Since $G$ is a Lie group, $i$ is continuous. The image $i(G_0)$ is connected because $G_0$ is connected. It contains $e$, since $i(e)=e^{-1}=e$. Therefore maximality of the connected component $G_0$ gives
\begin{align*}
i(G_0)\subset G_0.
\end{align*}
Thus $a^{-1}\in G_0$ for every $a\in G_0$.
Since $e\in G_0$, and $G_0$ is closed under multiplication and inversion, $G_0$ is a subgroup of $G$.
[/step]
[step:Conjugate the identity component by an arbitrary group element]
Fix $g\in G$. Define the conjugation map
\begin{align*}
c_g:G\to G,\qquad x\mapsto gxg^{-1}.
\end{align*}
This map is continuous because it is the composition of the continuous maps $x\mapsto gx$ and $x\mapsto xg^{-1}$. Since $G_0$ is connected, the image $c_g(G_0)=gG_0g^{-1}$ is connected. Also,
\begin{align*}
c_g(e)=geg^{-1}=e.
\end{align*}
Hence $gG_0g^{-1}$ is a connected subset of $G$ containing $e$, so
\begin{align*}
gG_0g^{-1}\subset G_0.
\end{align*}
Applying the same containment with $g^{-1}$ in place of $g$ gives
\begin{align*}
g^{-1}G_0g\subset G_0.
\end{align*}
Multiplying this inclusion on the left by $g$ and on the right by $g^{-1}$ gives
\begin{align*}
G_0\subset gG_0g^{-1}.
\end{align*}
Therefore
\begin{align*}
gG_0g^{-1}=G_0.
\end{align*}
Since $g\in G$ was arbitrary, $G_0$ is normal in $G$.
[/step]
[step:Combine closedness, subgroup structure, and conjugation invariance]
The first step shows that $G_0$ is closed in $G$. The multiplication and inversion steps show that $G_0$ is a subgroup of $G$. The conjugation step shows that $gG_0g^{-1}=G_0$ for every $g\in G$, which is precisely normality. Therefore $G_0$ is a closed [normal subgroup](/page/Normal%20Subgroup) of $G$.
[/step]
