Examples of continuous functions that are not homotopic to each other. Let $f,g:X \to Y$ be continuous functions. Then, consider $F:X \times I \to Y$ defined as follows:$$F(x,t) = f(1-t)+gt$$
How does one check whether $F$ is continuous(w.r.t the product topology)?
Is there any other way to define a homotopy between two functions?
As an example, take $f,g :  \mathbb{R} \to \mathbb{R}$ to be $f(x) = x, g(x) = sin(x)$. Then how does one PROVE that the homotopy map as defined above is continuous?
Moreover, what are some examples of maps that are not homotopic?
 A: The identity map of the circle $\rm{id}:S^1\to S^1$ and a constant map on $S^1$ are continuous but not homotopic.  This is due to a result by Heinz Hopf, from one point of view.
Intuitively the unit circle is not contractible.
By the alluded to result, we can wrap the circle around itself as many times as we like, continuously, and get non-homotopic maps.

The homotopy you refer to is the straight line homotopy.  There are plenty of others.
A: Generally, this all comes down to a space being path connected. If the path the homotopy described is contained in the space then it will be continous. For example, if we want to see the circular sector of radii 1 and 2 is homotopic to the unit circumference, we might consider the following map:
$$
f(x)=\frac{x}{||x||}
$$
It normalizes every point of the sector. It is continous because $0$ is not contained in the sector, thus, it is a retraction into the unit sphere. We want to prove it is a deformation retract, so we can end our proof, concluding that the two sets have the same type of homotopy. We consider the following homotopy:
$$
H(x,t)=f(x)·t+x·(1-t)
$$
Because of what I said earlier, one can check it is radial, making it thus continious (the sector is path connected).
A: The homotopy you describe for maps to $\mathbb{R}$ (or maybe $\mathbb{R}^n$) is continuous because addition and multiplication on $\mathbb{R}$ are continuous operations, along with the functions $f$ and $g$ being continuous.
In general, maps $f$ and $g$ are not homotopic if some $f(x)$ and $g(x)$ are in different path components (essentially using that if $f$ and $g$ were homotopic then so would be the compositions $\{x\} \hookrightarrow X \to Y$). This tells you for example that the identity of $\mathbb{R} - 0$ is not homotopic to the map $x \mapsto -x$.
Looking at such compositions with maps $Z \to X$ from other 'test' spaces $Z$ can often detect if $f$ and $g$ are not homotopic. For example looking at all maps $S^n \to X$ composed with $f$ and $g$ is the same as looking at the maps $\pi_n(X) \to \pi_n(Y)$ induced by $f$ and $g$.
