There is a cute argument for $b=c+1$ though I don't know how to pull it through in the general case.
Let $U$ be the set of ordered $c+1$-tuples of sets of cardinality $c$ in which no set repeats. Then, obviously, $|U|=(c+1)!{{n\choose c}\choose c+1}$. Let $V$ be the set of ordered $c$-tuples of sets of cardinality $c+1$ in which no set repeats. Then $|V|=c!{{n\choose c+1}\choose c}$. Now take any element of $V$, i.e., an ordered family of sets $A_1,\dots,A_{c}$. Take any $a_1\in A_1$ ($c+1$ choices). Suppose that $a_1\in A_1,\dots,a_k\in A_k$ are already chosen. Then we want to choose $a_{k+1}\in A_{k+1}$ so that $a_{k+1}\ne a_j$ and $A_{k+1}\setminus \{a_{k+1}\}\ne A_j\setminus\{a_j\}$ for all $j=1,\dots,k$. Suppose that we have $\ell$ prohibitions of the first type. Then, if we honor them, the prohibitions of the second type for the corresponding sets will be honored automatically (if $a_j\in A_{k+1}$ and we don't remove it, then surely $A_{k+1}\setminus \{a_{k+1}\}\ne A_j\setminus\{a_j\}$. The remaining $k-\ell$ sets introduce at most one prohibition of the second type each, so we have $\le k$ prohibitions total. Thus we have $\ge c+1-k$ choices for $a_{k+1}$. Once we run the whole procedure, the reduced sets $A_j$ listed in the same order and the set $\{a_1,\dots,a_c\}$ listed last will constitute an element of $U$ and each element of $V$ will generate $\ge(c+1)!$ different elements of $U$ (by different choices of $a_1,\dots,a_c$. On the other hand, each element of $U$ can be obtained from at most $c!$ elements of $V$ (that is the number of ways to distribute the elements of the last $c+1$-st set among the first $c$. This immediately gives a non-strict inequality you want. To make it strict, you need to assume $c>1$ and $n\ge c+1$ (otherwise you have equality), in which case the ordered sequence of sets $A_j=\{1,2,\dots,c+1\}\setminus\{j\}$ has fewer than $c!$ distribution options (none at all, really) resulting in a legitimate element of $V$.
Edit (the general case)
In this case it will be convenient to define $U$ as the set of all ordered $b$-tuples of sets of cardinality $c$ such that the sets are pairwise different (as sets), the first $c$ sets are unordered, and the last $b-c$ sets are ordered. Then $|U|=b!(c!)^{b-c}{{n\choose c}\choose b}$. Let $V$ be the same as before (ordered $c$-tuples of unordered subsets of cardinality $b$ in which the sets are pairwise different), so $|V|=c!{{n\choose b}\choose c}$.
Now let $\langle A_1,\dots, A_c\rangle$ be an element of $V$. We want to remove $b-c$ elements $a_{j,1},a_{j,2},\dots,a_{j,b-c}$ from $A_j$ and form $b-c$ new ordered sets $B_k=\langle a_{1,k},a_{2,k},\dots, a_{c,k}\rangle$ ($k=1,\dots,b-c$) to get an element of $U$.
To this end start with choosing $a_{1,1},\dots,a_{1,b-c}\in A_1$ in an arbitrary way, which gives $b(b-1)\dots(b-c+1)$ choices. Now, when choosing $a_{k+1,1}\in A_{k+1}$ for $k\ge 1$, add to the standard prohibitions $a_{k+1,1}\ne a_{j,1}$, $A_{k+1}\setminus\{a_{k+1,1}\}\ne A_j\setminus\{a_{j,1}\}$ ($j\le k$), which exclude at most $k$ elements just as before, the prohibitions $a_{k+1,1}\ne a_{1,m}$ ($m>1$), which exclude additional $b-c-1$ elements at most, leaving $c+1-k$ choices as before. These additional prohibitions guarantee that the set $B_1=\{a_{1,1},a_{2,1},\dots, a_{c,1}\}$ does not contain any of the elements $a_{1,m}$ ($m>1$) and, therefore, will be different from every set $B_m$ ($m>1$) as an unordered set and we still have $c!$ choices.
Having constructed $B_1$, we construct $B_2$ with the standard restrictions $a_{k+1,2}\ne a_{j,2}$, $A_{k+1}\setminus\{a_{k+1,1},a_{k+1,2}\}\ne A_j\setminus\{a_{j,1},a_{j,2}\}$ ($j\le k$) and additional restrictions $a_{k+1,2}\ne a_{1,m}$ but now with $m>2$. This gives $c!$ choices again, and so on. Thus, from each element of $V$, we get $b(b-1)\dots(b-c+1) (c!)^{b-c}$ distinct elements of $U$. The recovery of an element of $V$ from an element of $U$ is now possible in just one way, if at all, which, again, gives a non-strict inequality. I leave it to you to figure out when and how it becomes strict in this case.
Best Answer
Let $A$ be a set with $n$ elements. The involution $S \mapsto A\setminus S$ shows that $\binom{n}{k} = \binom{n}{n-k}$ for $0 \leqslant k \leqslant n$, so we can restrict our attention to $k \leqslant m := \bigl\lfloor \frac{n}{2}\bigr\rfloor$.
Suppose we want to select a committee of $m$ people, among which $k$ shall represent the committee to the outside world from a group of $n$ people. We can
Using the symmetry of the binomial coefficients, it follows that
$$\binom{n}{m} = \frac{\binom{n-k}{m-k}}{\binom{m}{m-k}}\cdot \binom{n}{k}.\tag{1}$$
Now we note that for $0 \leqslant r \leqslant s \leqslant t$ we have
$$\binom{s}{r} \leqslant \binom{t}{r},\tag{2}$$
with equality if and only if $s = t$ or $r = 0$.
Since $n-k \geqslant n-m \geqslant m$, using the inequality $(2)$ in $(1)$ shows the desired
$$\binom{n}{m} \geqslant \binom{n}{k},$$
with equality if and only if $k = m$.