Why a catalyst is not consumed in a reaction when all steps are reversible?
By saying Catalysts are not consumed by reactions. is meant there is no stoichimetric ratio to reactants, consuming catalysts and forming from them catalytically inactive compound.
By other words, some of reaction steps regenerates the original form of a catalyst, consumed by a prior step, so the net consumption is negligible.
The transition between active states of a catalyst is not considered a catalyst consumption, as it does not affect its catalytic activity.
Catalysis means the catalyst is reversibly switching between its 2 – chemically different – active states, producing an alternative and faster reaction path. Each form catalyzes one of the mutually opposite direactions of the net reaction. It is like a kind of a ferry to move compounds between reactant and product shores, bypassing a faraway bridge. The ferry is not consumed if ends at the opposite shore.

Note that things get little complicated, if some of products or intermediates catalyse a part or whole net reaction. Not being particularly aware of the intermediate case, but for the product, the famous case is oxidation of oxalate by permanganate in acidic environment:
$$\ce{5(COO)2^2-(aq) + 2 MnO4-(aq) + 16 H3O+(aq) -&gt; \\ -&gt;10 CO2(g) + 2 Mn^2+(aq) + 24 H2O}$$
$\ce{Mn^2+(aq)}$ as the product acts as a catalyst, forming catalytic system $\ce{Mn^3+(aq)/Mn^2+(aq)}$ with a similar redox potential as $\ce{MnO4-(aq)/Mn^2+(aq)}$.
$$\ce{4 Mn^2+(aq) + MnO4-(aq) + 8 H3O+(aq) &lt;=&gt; 5 Mn^3+(aq) + 12 H2O}$$
$$\ce{(COO)2^2-(aq) + 2 Mn^3+(aq)  -&gt; 2 CO2(g) + 2 Mn^2+(aq) }$$
The effective result is an autocatalytic reaction. There is sometimes applied a small amount of Mn^{+II} salt to speed up the reaction, or it is waited until initial addition of $\ce{KMnO4}$ creates enough $\ce{Mn^2+}$.