Topology/continuous functions of $\mathbb{R}^n$ using paths.

Question

We know that for a function $f:\mathbb{R}^n\to\mathbb{R}$ to be continuous, it is not sufficient for itto be continuous with respect to each coordinate. I believethe most commom counter-example is the function $f(x,y)=\dfrac{xy}{x^2+y^2}$ for $(x,y)\neq 0$ and $f(0)=0$.

What if we assume continuity "with respesct to every direction"?

Question 1: Given $f:\mathbb{R}^n\to\mathbb{R}$, suppose that for every $x\in\mathbb{R}^n$ and every $v\in\mathbb{R}^n$, the function $t\in\mathbb{R}\mapsto f(x+tv)\in\mathbb{R}$ is continuous. Is $f$ necessarily continuous?

Instead of considering only linear paths, we could consider arbitrary continuous paths: Let $\mathcal{C}$ be the set of continuous paths $\lambda:[0,1]\to\mathbb{R}^n$ (in the usual topologies).

Question 2: Suppose that for every $\lambda\in\mathcal{C}$, the function $f\circ\lambda:[0,1]\to\mathbb{R}$ is continuous. Is $f$ continuous?

And, going just a little further, we can admit arbitrary codomains (topological spaces), and this can be stated in terms of strong/final topologies;

Question 3: Is the topology of $\mathbb{R}^n$ the strong/final topology induced by $\mathcal{C}$?

Question 1 came to my mind some time ago, and questions 2 and 3 are simple adaptations.

My initial ideas would be to use some sort of (local) compactness to try to prove the continuity of $f$, or maybe use Peano curves or something of the sort to find a counter example, but I have no idea how to proceed.

Answer 1

For any $N\in\Bbb N$, you can easily make up functions that are continuous on every curve $y=cx^k$ for $k\le N$ [including the line $x=0$] but discontinuous (at the origin) when we restrict to $y=x^{N+1}$. I leave it to you to even construct one that is continuous along $y=cx^k$ for all $k\in\Bbb N$ but still discontinuous at the origin.

Nevertheless, the answer to question 2 is YES. I suggest you try to prove it by proving the contrapositive.

And the answer to question 3 is, if I understand you correctly, also YES. The strong topology will be the product topology, which is the standard metric topology on $\Bbb R^n$.