One possible definition of continuity is the following:
"A function $f:X\rightarrow Y$, where $X$ and $Y$ are topological spaces, is said to be continuous if for every open subset of $Y$, it's preimage by $f$ is open in $X$, i.e., if $V\subseteq Y$ is open (in $Y$), then $f^{-1}(V)$ is open (in $X$)."
In your argument, you only proved that the preimage of a specific open set in $Y$ ($Y$ itself) is open in $A$, which is not sufficient.
Now, let $f:X\rightarrow Y$ be a continuous function and $A\subseteq X$.
In the definition of subspace, a subset $V\subseteq A$ is open in $A$ iff there exists $U\subseteq X$ open in $X$ such that $V=U\cap A$.
Consider the restriction $f|_A:x\in A\mapsto f(x)\in Y$, and let $B$ be an open subset of $Y$. It should be clear that $f|_A^{-1}(B)=f^{-1}(B)\cap A$. Since $f$ is continuous from $X$ to $Y$ and $B$ is open in $Y$ then $f^{-1}(B)$ is open in $X$, hence $f|_A^{-1}(B)$ is the intersection of $A$ with an open subset of $X$, and is therefore open in $A$ (for every open subset of $B$).
We then conclude that $f|_A$ is a continuous function from $A$ to $Y$.
Best Answer
The distinction is that normal continuity of a map $f: X \to Y$ cannot be used to determine that a subset $U$ of $Y$ is open. Continuity gives the implication that $f^{-1}(U)$ is open provided $U$ is open, and being a quotient map means that you can additionally infer that $U$ is open provided that $f^{-1}(U)$ is open.
As an example, consider the map $f_{0}: \Bbb R \to \Bbb R$ sending all points to 0. Note that constant maps are always continuous. Now $f^{-1}(\{0\})$ is open, being all of $\Bbb R$. However $\{0\}$ is not open, demonstrating that $f_0$ is continuous but does not satisfy this stricter condition.
edit: Thanks to Harry Altman in the comments for noting that I was ignoring the surjectivity requirement. For a surjective example, consider the identity map $X_{\textrm{discrete}}\to X_{\textrm{indiscrete}}$ where the $X_{\textrm{discrete}}$ and $X_{\textrm{indiscrete}}$ is the discrete and indeiscrete topologies on an arbitrary set $X$. Here $f$ is continuous (follows from either the domain being discrete or the codomain being indiscrete) but for any nonempty proper $U \subset X$ we have that $U$ is not open in $X_{\textrm{indiscrete}}$ but $f^{-1}(U)$ is open in $X_{\textrm{discrete}}$.