I apologise if this seems too trivial which is why I am a bit suspicious.
I essentially need to prove that given ideals $I, J$ in a commutative ring $A$, $I \subset J$ iff $I_m \subset J_m$ in $A_m$, for all maximal ideals $m$ in $A$.
I proceeded as thus : Supposing $I \subset J $, take $\frac{x}{s} \in I_m$, where $x \in I$, $\implies x \in J$, $\implies \frac{x}{s} \in J_m$.
Conversely if $I_m \subset J_m$ in $A_m$, then take $x \in I$, $\implies \frac{x}{s} \in I_m, \implies \frac{x}{s} \in J_m$ and consequently $x \in J$.
Is this all there is to it??Or are there any loopholes that I havent been able to perceive??
The mistake in the OP's argument was discussed in the comments. Let give a proof (which differs from the one proposed in the comments which I think needs some finiteness-assumptions on the ideals; I was assuming $A$ is noetherian, I guess).
Given ideals $I,J\subset A$ such that $I_m \subset J_m\subset A_m$ for each maximal ideal $m\subset A$ we claim that $I\subset J$.
Let $f\in A$ be an arbitrary element and consider the ideal $(J:f) = \{a\in A\mid af\in J\}$. If $S\subset A$ is a multiplicative subset, then $f\in S^{-1}J$ holds if and only if $S\cap(J:f)\neq\emptyset$. In particular, if $m\subset A$ is a maximal ideal, then $f\in J_m$ if and only if $(J:f)$ is not(!) contained in $m$. Consequently, if $f\in J_m$ for each maximal ideal, then $(J:f)$ is not contained in any maximal ideal which is possible only if $(J:f) = A$; equivalently, if $f\in J$. Applied to each element of $I$ separately this implies that if $I_m\subset J_m$ for each maximal ideal $m\subset A$, then $I\subset J$.