Why do I need to specify some "continuous" choice of paths Let $X$ be some topological space and $Y$ be a path connected space.
If I have fixed $f: X\to Y$ continuous map and $y_0\in Y$. Then, I can find a path between $\phi_x: I \to Y$ between $f(x)$ and $y_0$. Thus, intuitively, I should be able to form a homotopy via $F: X\times I \to Y$ by the map $(x,t)\mapsto \phi_x(t)$. I have found another stack exchange post which says that this does not work because I have yet to define a "Continuous" choice of paths. What does this mean and where does my proof ($Y$ path connected -> $|[X,Y]|=1$) go wrong exactly?
 A: Your function $(x,t)\mapsto \phi_x(t)$ need not be continuous. And you need it to be continuous in order to do the homotopy theory.
For example consider $X=Y=\mathbb{R}$ and $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=x$ and fix $y_0=0$.
Now define $\phi_x:I\to\mathbb{R}$ as follows: $\phi_x(t)=tx$ except for $x=1$ where I define $\phi_1(t)=t^2$. Obviously each $\phi_x$ is a continuous path connecting $0$ to $f(x)$. I'm sure you can feel at this moment that this special $\phi_1$ will break the continuity of the induced homotopy. :)
Lets have a look at the induced homotopy:
$$H:\mathbb{R}\times I\to\mathbb{R}$$
$$H(x,t)=\phi_x(t)$$

Fact. $H$ is not continuous.

Proof. Let's take a sequence $v_n=(1-1/n, 1/2)$. Then
$$H(v_n)=H(1-1/n, 1/2)=\phi_{1-1/n}(1/2)=(1-1/n)\cdot 1/2$$
because $1-1/n\neq 1$. Thus
$$\lim H(v_n)=1/2$$
On the other hand
$$\lim v_n=(1, 1/2)$$
$$H(\lim v_n)=H(1, 1/2)=\phi_1(1/2)=1/4$$
So $H(\lim v_n)\neq \lim H(v_n)$ and this shows that $H$ is not continuous even though each $\phi_x$ is. $\Box$
So as you can see an arbitrary choice of $\phi_x$ does not work. You need a continuous choice of $\phi_x$ in the sense that $(x,t)\mapsto \phi_x(t)$ has to be a continuous function (in the case above it would be $\phi_x(t)=tx$ for all $x$). Only then you can conclude that there are non-homotopic functions (depending on $X,Y$).
