Consider a closed disc $D^n\subseteq \mathbb{R}^n$ and let $f:D^n\rightarrow \mathbb{R}^n$ be a continuous map with $f|_{S^{n-1}}$ is the identity on the boundary of $D^n$. Prove that $D^n \subseteq \operatorname{Im}(f)$. Is my following argument correct?
Suppose $x\in D^n$ and $x\not \in f(D^n)$. Then we have $\operatorname{id}_{S^{n-1}}$ factors through $f\circ i:S^{n-1} \hookrightarrow D^n \rightarrow \mathbb{R}^n-\{x\}$. Recalling the fact that $\mathbb{R}^n-\{x\}$ deform retracts to $S^{n-1}$, the contradiction here comes from applying the $H_{n-1}$ functor to obtain that $H_{n-1}(f\circ i):\mathbb{Z}\rightarrow 0 \rightarrow \mathbb{Z}$ is the zero homomorphism, certainly not equal to the identity homomorphism $H_{n-1}(\operatorname{id}_{S^{n-1}}):\mathbb{Z}\rightarrow \mathbb{Z}$. Hence $f$ must contain its image.
Also is there a way to prove this directly? I think I can smell Brouwer's fixed point theorem and/or Borsuk-Ulam but I'm not sure how to employ it.
Your proof is correct. You show that if $D^n$ is not contained in the image of $f$, then we would get a retraction $\phi : D^n \to S^{n-1}$ which would show that $id: S^{n-1} \to S^{n-1}$ is null-homotopic. There are various approaches to falsify this, you do it via homology.
Brouwers's fixed point theorem is usually proved via homology (using the non-existence of a retraction $\phi : D^n \to S^{n-1}$), but there are also other proofs. See for example Why is Brouwer’s Fixed Point Theorem considered a result of algebraic topology?
Anyway, if you take Brouwers's fixed point theorem as a given result, then you can easily show that $S^{n-1}$ is not a retract of $D^n$: Assume there exists a retraction $\phi : D^n \to S^{n-1}$. Then $\psi : D^n \to S^{n-1}, \psi(x) = -\phi(x)$, is well-defined and does not have a fixed point. In fact, no $x$ in the open unit ball can be a fixed point, and no $x \in S^{n-1}$ can be a fixed point because $\psi(x) = - \phi(x) = -x \ne x$.