[proofplan]
The proof uses only the compatibility between the elliptic curve group law and the topology on the real locus. First we record that addition and inversion on $E(\mathbb{R})$ are continuous. Then we use elementary connectedness principles: products and continuous images of connected sets are connected, and any connected subset containing $O$ lies in the connected component of $O$. Applying these principles to addition and inversion gives closure under the group operation and under inverses.
[/proofplan]
[step:Record the continuity of the elliptic curve group operations]
Let $m: E(\mathbb{R}) \times E(\mathbb{R}) \to E(\mathbb{R})$ denote the addition map defined by $m(P,Q) := P+Q$ for $(P,Q) \in E(\mathbb{R}) \times E(\mathbb{R})$. Let $\iota: E(\mathbb{R}) \to E(\mathbb{R})$ denote the inversion map defined by $\iota(P) := -P$ for $P \in E(\mathbb{R})$.
For a nonsingular real elliptic curve, addition and inversion are morphisms of real algebraic varieties. Morphisms are continuous on real points with respect to the Euclidean topology, because in affine charts they are given by regular functions, and regular functions are continuous. Therefore the induced maps $m$ and $\iota$ on the real locus are continuous.
[/step]
[step:Show that sums of points in the identity component remain in the identity component]
Let $C := E(\mathbb{R})^0$, so $C$ is the connected component of $E(\mathbb{R})$ containing $O$.
We first note that $C \times C$ is connected. Indeed, if $C \times C$ were separated as $A \cup B$ by two nonempty disjoint relatively open subsets, then each horizontal slice $C \times \{Q\}$ and each vertical slice $\{P\} \times C$ would be connected and hence contained in one side of the separation. Since any two points $(P_1,Q_1)$ and $(P_2,Q_2)$ can be joined inside the union
\begin{align*}
(C \times \{Q_1\}) \cup (\{P_2\} \times C),
\end{align*}
the separation cannot place them in different sides. Thus no such separation exists.
Since $m$ is continuous, the image $m(C \times C)$ is connected. Moreover,
\begin{align*}
m(O,O) = O + O = O,
\end{align*}
so $m(C \times C)$ is a connected subset of $E(\mathbb{R})$ containing $O$. By maximality of the connected component $C$, we have
\begin{align*}
m(C \times C) \subset C.
\end{align*}
Thus, for every $P,Q \in C$, one has $P+Q \in C$.
[guided]
The goal of this step is to prove closure under addition. We cannot argue pointwise from the definition of a connected component; instead, we package all possible sums into one connected set.
Let $C := E(\mathbb{R})^0$. By definition, $C$ is the maximal connected subset of $E(\mathbb{R})$ containing the identity element $O$. Consider the product set $C \times C \subset E(\mathbb{R}) \times E(\mathbb{R})$. We claim that $C \times C$ is connected. To justify this, suppose for contradiction that $C \times C = A \cup B$ is a separation by nonempty disjoint relatively open sets. For each fixed $Q \in C$, the horizontal slice $C \times \{Q\}$ is homeomorphic to $C$, hence connected. Therefore it must lie entirely in $A$ or entirely in $B$. Similarly, for each fixed $P \in C$, the vertical slice $\{P\} \times C$ is connected and must lie entirely in one side of the separation.
Now take two arbitrary points $(P_1,Q_1),(P_2,Q_2) \in C \times C$. The set
\begin{align*}
(C \times \{Q_1\}) \cup (\{P_2\} \times C)
\end{align*}
is connected because it is the union of two connected sets with common point $(P_2,Q_1)$. This connected set contains both $(P_1,Q_1)$ and $(P_2,Q_2)$, so those two points cannot lie in different sides of a separation. Since the two points were arbitrary, the proposed separation is impossible. Hence $C \times C$ is connected.
The addition map $m: E(\mathbb{R}) \times E(\mathbb{R}) \to E(\mathbb{R})$, defined by $m(P,Q) := P+Q$, is continuous, so its image $m(C \times C)$ is connected. This image contains $O$, because
\begin{align*}
m(O,O) = O+O = O.
\end{align*}
Since $C$ is the connected component containing $O$, every connected subset of $E(\mathbb{R})$ that contains $O$ must be contained in $C$. Therefore
\begin{align*}
m(C \times C) \subset C.
\end{align*}
Equivalently, if $P,Q \in C$, then $P+Q \in C$. This proves closure of $C$ under addition.
[/guided]
[/step]
[step:Show that inverses of points in the identity component remain in the identity component]
Since $\iota$ is continuous and $C$ is connected, the image $\iota(C)$ is connected. Also,
\begin{align*}
\iota(O) = -O = O,
\end{align*}
so $\iota(C)$ is a connected subset of $E(\mathbb{R})$ containing $O$. By maximality of the connected component $C$, we have
\begin{align*}
\iota(C) \subset C.
\end{align*}
Thus, for every $P \in C$, one has $-P \in C$.
[/step]
[step:Conclude that the identity component is a subgroup]
The set $C$ contains $O$ by definition. The preceding steps show that $C$ is closed under addition and under taking inverses. Therefore $C$ is a subgroup of $E(\mathbb{R})$. Since $C = E(\mathbb{R})^0$, this proves that $E(\mathbb{R})^0$ is a subgroup of $E(\mathbb{R})$.
[/step]
