Since $A \subset B$, whenever $A$ occurs, $B$ also occurs. Hence $A \cap B = A.$

By definition of conditional probability, $P(A\mid B)=\frac{P(A\cap B)}{P(B)}.$ Using $A\cap B=A$, $P(A\mid B)=\frac{P(A)}{P(B)}.$

Now compare this with the given options.

Option A: $P(A\mid B)=1$

This would mean $\frac{P(A)}{P(B)}=1 \implies P(A)=P(B),$ which is not always true. So A is false.

Option B: $P(A\mid B)=P(B)-P(A)$

But actually, $P(A\mid B)=\frac{P(A)}{P(B)},$ which is not generally equal to $P(B)-P(A)$. So B is false.

Option C: $P(A\mid B)\le P(A)$

Since $P(A\mid B)=\frac{P(A)}{P(B)},$ and because $0<P(B)\le 1$ (for conditional probability to be defined, $P(B)>0$), dividing by $P(B)$ gives $\frac{P(A)}{P(B)}\ge P(A).$ So C is false.

Option D: $P(A\mid B)\ge P(A)$

From above, $P(A\mid B)=\frac{P(A)}{P(B)}\ge P(A),$ because $P(B)\le 1$ and $P(B)>0$. Hence D is always true.

Therefore, the correct answer is $\boxed{D}.$