The correct observation in the following reaction is:
$$Sucrose \xrightarrow[Cleavage \; (Hydrolysis)]{Glycosidic \; bond \; } A + B \xrightarrow[ Reagent]{Seliwanoff } ?$$

We have the disaccharide $$\text{Sucrose}$$, which consists of one unit of $$\alpha\!-\!D\!-\!\text{glucopyranose}$$ and one unit of $$\beta\!-\!D\!-\!\text{fructofuranose}$$ linked through a glycosidic bond $$\left(C\_1^{\text{Glucose}}{-}O{-}C\_2^{\text{Fructose}}\right)$$.

According to basic carbohydrate chemistry, “glycosidic bond cleavage (hydrolysis)” breaks this bond in the presence of dilute acid or the enzyme invertase. Stating the hydrolysis process explicitly:

Sucrose} + H_2O \;\xrightarrow[\text{dil.\;HCl / invertase}]{\text{Hydrolysis}}\; \text{Glucose} + \text{Fructose

So after the first step we obtain two monosaccharides:

$$A = \text{Glucose} \quad\text{and}\quad B = \text{Fructose}$$

Now we treat this mixture with Seliwanoff reagent. First, let us recall what the Seliwanoff test tells us:

The reagent is a solution of $$\text{resorcinol}$$ in concentrated $$HCl$$. Under strong acid, any ketose (but not an aldose) is rapidly dehydrated to form $$5\text{-hydroxymethylfurfural}$$. This intermediate then condenses with resorcinol to yield a cherry-red coloured complex. Aldoses either react very slowly or give no such colour within the usual test time.

In our hydrolysate mixture,

$$\text{Glucose} \;(\text{aldohexose, no rapid colour})$$

$$\text{Fructose} \;($$ ketohexose, gives red colour rapidly $$)$$

Because fructose is a ketohexose, it is the one that reacts instantly with the Seliwanoff reagent, producing the characteristic red colour. The presence of glucose does not mask this observation; it merely lags behind, so the cherry-red colour of the fructose dominates.

Therefore, the correct experimental observation is the formation of a red colour.

Hence, the correct answer is Option C.