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I am trying to prove the following theorem.

Let $\gamma=\gamma(s)$ be a regular space curve, parameterised by its arc-length $s$. If $\kappa=0$ is the curvature of $\gamma$, then $\gamma(s)$ is a straight line.

Now I proceeded to use the first Serret-Frenet equation $\mathbf t'=\kappa \mathbf n$ which gives us $\mathbf t'=\mathbf 0$, in other words, $\mathbf t$ is a constant vector. Clearly if a curve has a constant tangent vector, then it must be a straight line.

When I looked up my lecturer's proof, he proved it similarly but performed an additional step:

Now $$\frac{\mathrm d}{\mathrm d s}(\mathbf r\times\mathbf t)=\mathbf t\times\mathbf t+\mathbf r\times\mathbf t'=\mathbf 0,$$ so $\mathbf r\times\mathbf t$ is also a constant vector, i.e. $\mathbf r\times\mathbf t=\mathbf a$ where $\mathbf a$ is some constant vector. This corresponds to the equation of a straight line.

It is this step which I do not understand. How does this correspond to the equation of a straight line? Shouldn't that be $\mathbf r=\mathbf a+t\mathbf b$? And is this additional step necessary to prove the result?

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    Well, $\boldsymbol{\gamma}''(s)=\mathbf{t}'(s)=0$ is sufficient to for a differentiable curve in euclidean space to be a straight line. So I'm not sure about why that extra step is done. On the other hand for a circle $\boldsymbol{\gamma}(s)=\begin{pmatrix}\cos(s)\\ \sin(s)\\0\end{pmatrix}$ you have that $\boldsymbol{\gamma}(s)\times\mathbf t(s)=\begin{pmatrix}0\\0\\1\end{pmatrix}$ being constant, so that cross product being constant is not the equation of a line. – 2017-01-23
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    @s.harp -- The vector $\vec t$ is not constant for a circle. The equation $\vec r \times \vec t = \vec a$, where both $\vec t$ (non-zero) and $\vec a$ (orthogonal to $\vec t$) are constant, does describe a line. – 2018-01-15

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I'm with you, although your sentence starting with "Clearly" could use an explicit proof. Namely, from $\mathbf t=\mathbf c$ constant, you should integrate to get $\boldsymbol\gamma(s)=s\mathbf c + \mathbf d$.

All I see on first principles from your lecturer's proof is that $\mathbf r$ lies in the plane through the origin with normal vector $\mathbf a$.