How to prove a statement by mathematical induction?

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying $f(x+y) = f(x) + f(y)$ for any $x, y$ in $\mathbb{R}$.

Suppose $f(x)$ is continuous at 0, and $f(1)$ equals a constant $a$. Show that $f(x) = ax$

The hint provided for this question is to first show that $f(r) = ar$ for $r$ in $\mathbb{Q}$. So I want to express $f(1)$ in terms of $p$ and $q$ and $f(r) = f(p/q)$, where $q$ is in $\mathbb{N}$ and $p$ is in $\mathbb{Z}$. To this end, I want to show that $f(1) = qf(1/q)$ by mathematical induction. I am stuck at the step after the inductive hypothesis.

For $q=1$, we have $1f(1/1) = f(1)$, so $f(1) = qf(1/q)$ holds.
Inductive hypothesis: Assume $kf(1/k) = f(1)$ holds for $q = k$ in $\mathbb{N}$.
For $q=k+1$, we have $(k+1)f(1/(k+1))$.

I am stuck at 3. I think I have to express $(k+1)f(1/(k+1))$ in terms of $k$ and $f(1/k)$ in order to complete the proof for this step, and I am unable to show something like that. I feel like I am missing something simple here.

Why not abandon the hint and instead show that $f(kx) = kf(x)$ for any $x \in \mathbb{R}$? That would be easier to complete the induction. — Yuki.F, May 14 '21 at 04:57
FYI, a few closely related questions are Properties of Cauchy Functions, Let $f$ be a real function such that, for $x,y\in \mathbb{R}$ $f(x+y)=f(x)+f(y)$ Prove that (a) $f(0)=0$ (b) $f(n)=nf(1)$ and Checking my understanding of Cauchy's functional equation $f(x+y)=f(x)+f(y)$. — John Omielan, May 14 '21 at 05:10
Since $f(x) + f(y) = f(x+y)$, for any $x, y \in \mathbb{R}$ you have that $f(nx) = nf(x)$ for any $n \in \mathbb{N}$ and any $x \in \mathbb{R}$ (this is easy to prove by induction). Then we have $qf(1/q) = f(q/q) = f(1)$ for any $q \in \mathbb{N}$. I don't really understand why do you need this to solve the problem but I hope it helps. — I. Roperval, May 14 '21 at 05:20

How to prove a statement by mathematical induction?

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