[proofplan]
We write $O$ for the point at infinity on $E$ and write $P \oplus Q$ for the elliptic curve group law; this is the addition operation denoted $P+Q$ in the theorem statement. We prove that every case in the chord-and-tangent definition of this group law preserves rational coordinates. The identity and vertical-line cases give $O$, which is rational by definition. In the remaining secant and tangent cases, the affine chord-and-tangent formulas define the third point of $E$ and express its coordinates using rational field operations with nonzero denominators.
[/proofplan]
[step:Reduce to the affine nonvertical cases]
Let $O \in E(\mathbb{Q})$ denote the point at infinity, and let $\oplus: E(\mathbb{Q}) \times E(\mathbb{Q}) \to E$ denote the chord-and-tangent addition operation on $E$. This operation is the elliptic curve addition written as $+$ in the theorem statement, so proving $P \oplus Q \in E(\mathbb{Q})$ proves $P+Q \in E(\mathbb{Q})$.
If $P = O$, then $P \oplus Q = Q \in E(\mathbb{Q})$ by the identity law in the chord-and-tangent definition. If $Q = O$, then $P \oplus Q = P \in E(\mathbb{Q})$ by the same identity law.
Assume now that neither point is $O$. Write
\begin{align*}
P = (x_1,y_1), \qquad Q = (x_2,y_2),
\end{align*}
where $x_1,y_1,x_2,y_2 \in \mathbb{Q}$ and
\begin{align*}
y_1^2 = x_1^3 + ax_1 + b, \qquad y_2^2 = x_2^3 + ax_2 + b.
\end{align*}
If $x_1 = x_2$ and $y_2 = -y_1$, then $Q = -P$ in the chord-and-tangent group law, the vertical-line case in the definition gives $P \oplus Q = O$, and $O \in E(\mathbb{Q})$ by definition.
It remains to handle the secant case $x_1 \neq x_2$ and the tangent case $P=Q$ with $y_1 \neq 0$.
[/step]
[step:Compute the secant sum using rational field operations]
Assume $x_1 \neq x_2$. Define the secant slope $\lambda \in \mathbb{Q}$ by
\begin{align*}
\lambda := \frac{y_2-y_1}{x_2-x_1}.
\end{align*}
The denominator is nonzero because $x_1 \neq x_2$, so $\lambda$ is rational. In the nonvertical secant case, the defining affine formula for the chord-and-tangent group law defines
\begin{align*}
x_3 := \lambda^2 - x_1 - x_2, \qquad
y_3 := \lambda(x_1-x_3)-y_1.
\end{align*}
Since $\mathbb{Q}$ is closed under addition, subtraction, multiplication, and division by nonzero rational numbers, we have $x_3,y_3 \in \mathbb{Q}$. In this case
\begin{align*}
P \oplus Q = (x_3,y_3),
\end{align*}
so $P \oplus Q \in E(\mathbb{Q})$.
[guided]
In the secant case, the only possible obstruction to rationality is division by the horizontal difference $x_2-x_1$. We are assuming $x_1 \neq x_2$, so this denominator is a nonzero rational number. Therefore the slope
\begin{align*}
\lambda := \frac{y_2-y_1}{x_2-x_1}
\end{align*}
belongs to $\mathbb{Q}$.
The defining affine chord-and-tangent formula in the nonvertical secant case then gives the third point of $E$ by
\begin{align*}
x_3 := \lambda^2 - x_1 - x_2, \qquad
y_3 := \lambda(x_1-x_3)-y_1.
\end{align*}
Each expression is formed from rational numbers by field operations in $\mathbb{Q}$: squaring, subtraction, multiplication, and addition. Hence $x_3,y_3 \in \mathbb{Q}$. Since the group law defines $P \oplus Q$ to be this affine point in the nonvertical secant case, we conclude
\begin{align*}
P \oplus Q = (x_3,y_3) \in E(\mathbb{Q}).
\end{align*}
[/guided]
[/step]
[step:Compute the tangent sum using rational field operations]
Assume $P=Q$ and $y_1 \neq 0$. Define the tangent slope $\lambda \in \mathbb{Q}$ by
\begin{align*}
\lambda := \frac{3x_1^2+a}{2y_1}.
\end{align*}
The denominator is nonzero because $y_1 \neq 0$, so $\lambda$ is rational. In the nonvertical tangent case, the defining affine formula for the chord-and-tangent group law defines the third point of $E$ by
\begin{align*}
x_3 := \lambda^2 - 2x_1, \qquad
y_3 := \lambda(x_1-x_3)-y_1.
\end{align*}
Again $x_3,y_3 \in \mathbb{Q}$ because they are obtained from rational numbers by rational field operations. Thus
\begin{align*}
P \oplus P = (x_3,y_3) \in E(\mathbb{Q}).
\end{align*}
[/step]
[step:Verify that the cases exhaust the addition law]
For affine points $P=(x_1,y_1)$ and $Q=(x_2,y_2)$, either $x_1 \neq x_2$ or $x_1=x_2$. The first alternative is the secant case handled above. In the second alternative, the curve equation gives
\begin{align*}
y_1^2 = x_1^3+ax_1+b = x_2^3+ax_2+b = y_2^2,
\end{align*}
so $y_2 = y_1$ or $y_2=-y_1$. If $y_2=-y_1$, the vertical-line case gives $P \oplus Q = O$. If $y_2=y_1$, then $P=Q$; when $y_1=0$ this is again the vertical-line case because $P=-P$, and when $y_1 \neq 0$ it is the tangent case handled above.
Thus every possible pair $P,Q \in E(\mathbb{Q})$ falls into one of the cases already proved. Therefore $P \oplus Q \in E(\mathbb{Q})$ for all $P,Q \in E(\mathbb{Q})$. Since $\oplus$ is the addition operation denoted by $+$ in the theorem statement, this proves $P+Q \in E(\mathbb{Q})$.
[/step]
