Condition for an additive function to be continuous

The problem below is Problem 7 from this year's Miklos Schweitzer competition (contest ended Nov 4th).

Suppose that $f: \Bbb{R} \to \Bbb{R}$ is an additive function (that is $f(x+y) = f(x)+f(y)$ for all $x, y \in \Bbb{R}$) for which $x \mapsto f(x)f(\sqrt{1-x^2})$ is bounded of some nonempty subinterval of $(0,1)$. Prove that $f$ is continuous.

I attempted a solution by contradiction, but didn't got very far. I was thinking to exploit the fact that a discontinuous additive function is unbounded on every interval and its graph is dense in $\Bbb{R}^2$.

First observe that for every $\epsilon > 0$ there exist rational numbers $p,q > 0$ with $p<\epsilon$ and $p^2 + q^2 = 1$. This is just a consequence of well-known formulas for construction of Pythagorean triples. E.g., we can choose $$ p = \frac{2n+1}{n^2 + (n+1)^2} \quad \text{ and } q = \frac{2n(n+1)}{n^2 + (n+1)^2} $$ for $n \in \mathbb{N}$ sufficiently large. For such pairs we define an associated rotation matrix $$ A_p = \begin{bmatrix} q & p \\ -p & q \end{bmatrix}. $$

The assumption in the problem means that the function $$ F(x,y) = f(x) f(y) $$ is bounded on some arc $I$ of the unit circle $x^2 + y^2 = 1$.

Now if $(x,y)$ is arbitrary and $A_p$ is a rational rotation matrix as above, then (identifying the matrix and the induced linear map) \begin{align*} F(A_p(x,y)) &= F(qx + py, -px + qy) = f(qx+py) f(-px+qy) \\ &= (qf(x)+ pf(y)) (-pf(x) + qf(y)) \\ &= qp(f(y)^2 - f(x)^2) + (q^2 - p^2) f(x)f(y), \end{align*} where we used the (easily proved) fact that $f$ is linear over the rational numbers.

Now we fix a small subarc $J$ of $I$, and a rational rotation matrix $A_p$ with $p>0$ such that $A_p(J) \subset I$. (This is possible because by the argument in the first paragraph we can find $A_p$ arbitrarily close to the identity.) For $(x,y) \in J$ the above calculation then implies that $$ f(y)^2 - f(x)^2 = \frac{F(A_p(x,y))}{qp} - \frac{(q^2 - p^2) F(x,y)}{qp} $$ so that $f(y)^2 - f(x)^2$ is bounded on $J$. Since $f(x)f(y)$ is also bounded on $J$, it follows that $(f(y)+if(x))^2 = f(y)^2 - f(x)^2 + 2i f(x)f(y)$ is bounded on $J$, so that $f$ is bounded on the projections of $J$ to both coordinate axes. As stated in the question, local boundedness of $f$ implies continuity.