I have an intriguing question. There's a proposition or exercise stating that if $X$ is a simply connected space, then the map $f: S^1 \longrightarrow X$ can be continuously extended to a disk $D^2$.
On the other hand, we know that if $f$ is null-homotopic, it can also be extended to $D^2$. Utilizing this insight, I will employ it to establish the previously mentioned proposition.
So to prove this proposition, it suffices to show that $f$ is null-homotopic. Let $(0,1)$ be a point in $S^1$, and consider the constant map $c_{f(0,1)}: S^1 \longrightarrow X$, defined as $c_{f(0,1)}(x,y) = f(0,1)$ for all $(x,y) \in S^1$.
Now, take a homotopy $H: S^1 \times [0,1] \longrightarrow X$, defined as $H((x,y), t) = f(tx, 1-t+ty)$. We observe that $H((x,y), 0) = f(0,1) = c_{f(0,1)}(x,y)$ and $H((x, y), 1) = f(x,y)$. Thus, $f$ is null-homotopic, and consequently, it is extended to $D^2$.
Surprisingly, the proof doesn't explicitly rely on the assumption that $X$ is simply connected. Is there something missing, or is there an inherent reason for this?
Your proof is incorrect, since the homotopy $H$ is not well-defined. You want to calculate $f(tx,1-t+ty)$ for arbitrary $t\in[0,1]$ and $(x,y)\in S^1$. But $f$ is only defined on $S^1$. If $t\neq0,1$ and $y\neq1$ we have $(tx,1-t+ty)\notin S^1$, hence you cannot calculate $f(tx,1-t+ty)$.
Hint for a correct approach: A continuous map $f\colon S^1\to X$ can be interpreted as the element of $\pi_1(X,f(0,1))$ given by $$ \begin{array}{rcl} \tilde f\colon[0,1]&\to& X\\ t&\mapsto& f(\sin(2\pi\operatorname i\cdot t),\cos(2\pi\operatorname i\cdot t)) \end{array} $$ If $\pi_1(X,f(0,1))=0$ there exists a path-homotopy $\tilde H\colon[0,1]\times[0,1]\to X$ from $\tilde f$ to the constant path at $f(0,1)$. One can upgrade this path-homotopy to a homotopy $H\colon S^1\times[0,1]\to X$ that shows that $f$ is null-homotopic.