To provide a bit of a wider context on ulo's answer, consider the following general situation. Suppose $g,g':A\to B$ and $f,f':B\to C$ are maps with $g\simeq g'$ and $f\simeq f'$ via homotopies $H:A\times [0,1]\to B$ and $I:B\times[0,1]\to C$. Then $f\circ g$ is homotopic to $f'\circ g'$, by just "composing" the two homotopies. To be precise, you use the homotopy $J:A\times [0,1]\to C$ defined by $J(a,t)=I(H(a,t),t))$.
In your case, you can take $A=B=X$, $C=Y$, $g=\mathrm{id}_X$, $g'=c$, and $f=f'$ to be your arbitrary map $X\to Y$. You get that $f\circ g=f\circ\mathrm{id}_X=f$ is homotopic to $f'\circ g'=f\circ c$, and $f\circ c$ is a constant map because $c$ is a constant map.
Since $\mathbb{R}^2 - (0, 0)$ is path connected, there exists a continuous path $h: [0, 1] \rightarrow \mathbb{R} - (0, 0)$ such that $h(0) = f(0)$ and $h(1) = g(0)$.
Define $F:[0, 1]\times [0, 1] \rightarrow \mathbb{R}^{2}-\{(0,0)\}$ as
$$
F(x, t) =
\begin{cases}
f((1-3t)x), & t \in [0, \frac{1}{3}] \\
h(3t-1), & t \in [\frac{1}{3}, \frac{2}{3}] \\
g((3t-2)x), & t \in [\frac{2}{3}, 1]
\end{cases}
$$
where the three cases are continuous and agree at the two overlapping points $t=\frac{1}{3}$ and $t=\frac{2}{3}$. Therefore, $F$ is continuous. Moreover, $F(x, 0) = f(x)$ and $F(x, 1) = g(x)$ for all $x \in \mathbb{R}^2 - (0, 0)$. Thus, $F$ is a desired homotopy between $f$ and $g$.
Note that we cannot define $F$ using $F(x,t)=(1-t)f(x)+tg(x)$ since there may exist $x$ and $t$ such that $F(x, t) = (0, 0)$. Above, we avoid this problem since all three functions $f$, $g$ and $h$ map into $\mathbb{R}^2 - (0, 0)$.
Best Answer
A path $y_0 \rightarrow y_1$ is the same as a homotopy between the maps $x_0 \rightarrow y_0$ and $x_0 \rightarrow y_1$ since $I \times \{x_0 \} \approx I$. So if you have two maps $f,g:X \rightarrow Y$ with the properties you mention you precompose them with any inclusion $i:\{x_0 \} \rightarrow X$ to get the contradiction. If $f \simeq g$ then $f \circ i \simeq g \circ i$ which would mean that there exists a path between $y_0$ and $y_1$.