For the sake of completeness here is a slightly different approach to part one, which then continues along the path in the link from Brian Scott.
The generating function of the set of partitions by the number of parts is given by
$$ G(z,u) = \prod_{k\ge 1} \frac{1}{1-uz^k}.$$
It follows that the generating function of the set of partitions with an even number of parts is given by
$$ G_1(z) = \frac{1}{2} G(z, 1) + \frac{1}{2} G(z, -1)
= \frac{1}{2} \prod_{k\ge 1} \frac{1}{1-z^k}
+ \frac{1}{2} \prod_{k\ge 1} \frac{1}{1+z^k}.$$
Similarly for an odd number of parts,
$$ G_2(z) = \frac{1}{2} G(z, 1) - \frac{1}{2} G(z, -1)
= \frac{1}{2} \prod_{k\ge 1} \frac{1}{1-z^k}
- \frac{1}{2} \prod_{k\ge 1} \frac{1}{1+z^k}.$$
Therefore
$$ G_1(z) - G_2(z) = Q(z) = \prod_{k\ge 1} \frac{1}{1+z^k}$$
and this is the generating function of the difference between the number of partitions into an even and odd number of parts.
Now observe that this is
$$\prod_{k\ge 1} \frac{1-z^k}{1-z^{2k}}
= \prod_{k\ge 0} (1-z^{2k+1})$$
because the denominator cancels all the factors with even powers in the numerator.
Here is an important observation: the above expression of $Q(z)$ enumerates partitions into unique odd parts in a generating function of signed coefficients where the sign indicates the parity of the number of parts. There is no cancellation between partitions that add up to the same value because the counts of constituent parts have the same parity. Moreover as all parts are odd, partitions of odd numbers must have an odd number of parts and of even numbers, an even number. Therefore the coefficients of $Q(z)$ alternate in sign.
To get the series that generates the absolute values of these coefficients, we create generating functions for the even powers and the odd ones, inverting the sign of the coefficients of the odd ones. The even ones are generated by
$$\frac{1}{2} Q(z) + \frac{1}{2} Q(-z)$$
and the odd ones by
$$-\left(\frac{1}{2} Q(z) - \frac{1}{2} Q(-z)\right)$$
Adding these gives $$ Q(-z).$$
But this is
$$\prod_{k\ge 0} (1-(-z)^{2k+1})
= \prod_{k\ge 0} (1-(-1)^{2k+1} z^{2k+1})
= \prod_{k\ge 0} (1+ z^{2k+1}),$$
precisely the generating function of partitions into distinct odd parts.
This is sequence A000700 from the OEIS.
In the "Update" (the day after the Question itself was posted), the OP mentions a recursion for counting the partitions of $n$ into $k$ distinct parts, each part at most $M$:
$$ p_k(\leq M, \mathcal{D},n) =
p_{k-1}(\leq M-1, \mathcal{D},n-k) +
p_k(\leq M-1, \mathcal{D},n-k) $$
and asks "How can I prove this?".
To see this, separate the required partitions on the basis of whether $1$ appears as a summand. If it does, then subtracting $1$ from each summand produces a partition of $n-k$ with exactly $k-1$ distinct parts (since the original summand $1$ disappears), each part at most $M-1$. These partitions are counted by the first term on the right-hand side of the recursion. Otherwise the summand $1$ does not appear, and subtracting $1$ from each part resulting in a partition of $n-k$ with exactly $k$ distinct parts, each part at most $M-1$. These cases are counted by the second term.
Note that a partition of $n$ with $k$ distinct parts exists if and only if $n \ge \binom{k+1}{2}$, because the ascending summands $m_1 + \ldots + m_k = n$ must satisfy $m_i \ge i$. If $n = \binom{k+1}{2}$, then there is just one such partition with $k$ distinct parts, the largest of which is $k$. Repeated application of the recursion will culminate with terms which we can evaluate "by inspection" as either zero or one.
By itself this recursion doesn't seem to give us an especially attractive way of evaluating $p_k(\leq M, \mathcal{D},n)$. Like the recursion for Fibonacci numbers, as a top-down method it suffers from recalculating terms multiple times (giving exponential complexity), so we would be better off working with it as a bottom-up method (giving polynomial complexity).
Better for large parameters is to close the circle with the ideas presented by @NikosM. by showing how the evaluation of $p_k(\leq M, \mathcal{D},n)$ can be reduced to counting restricted partitions without requiring distinct summands.
Prop. Suppose that $n \gt \binom{k+1}{2}$. Then the following are equal:
$(i)$ the number of partitions of $n$ into exactly $k$ distinct parts, each part at most $M$, i.e. $p_k(\leq M, \mathcal{D},n)$
$(ii)$ the number of partitions of $n - \binom{k}{2}$ into exactly $k$ parts, each part at most $M$
$(iii)$ the number of partitions of $n - \binom{k+1}{2}$ into at most $k$ parts, each part at most $M$
Sketch of proof: Once we know the ordered summands of partitions in $(i)$ satisfy $m_i \ge i$, it is easy to visualize their equivalents in $(ii)$ and $(iii)$ by Young tableaux, also called Ferrers diagrams. We remove a "base triangle" of dots corresponding to the first $i$ dots in the $i$th summand (since $m_i \ge i$) to get case $(iii)$, and remove one fewer dot in each summand to preserve exactly $k$ summands in case $(ii)$. These constructions are reversible, and the counts are equal.
Remark 1 If $n \le \binom{k+1}{2}$, $p_k(\leq M, \mathcal{D},n)=1$ if $n=\binom{k+1}{2}$ and $k \le M$ and otherwise $p_k(\leq M, \mathcal{D},n)=0$.
Remark 2 For fixed $k,n$, suppose $M_0 = n - \binom{k}{2} \gt 0$. Then for all $M \ge M_0$, $p_k(\leq M, \mathcal{D},n) = p_k(\leq M_0, \mathcal{D},n)$. That is, further increasing the upper bound $M$ on the size of parts will not yield additional partitions of $n$ with exactly $k$ distinct parts.
Recommended Reading
Andrews, George E. The Theory of Partitions (Cambridge University Press, 1998)
A modern classic for theory of integer partitions, reviewed by Richard Askey in BAMS.
Best Answer
Here the lazy person approach: the number of partitions is given by $$ p(n) = [x^n]\frac{1}{(1-x)(1-x^2)(1-x^3)\cdots} $$ while the number of partitions into distinct odd parts is given by $$ p_1(n) = [x^n](1+x)(1+x^3)(1+x^5)\cdots $$ so we just need to prove that all the coefficients of $(1-x)(1-x^2)(1-x^3)\cdots(1+x)(1+x^3)(1+x^5)\cdots$ except the very first one are even. On the other hand in $\mathbb{F}_2[[x]]$ we have $$ \prod_{n\geq 1}(1-x^n)\prod_{m\geq 0}(1+x^{2m+1})=\prod_{n\geq 1}(1-x^{2n})\prod_{m\geq 0}(1+x^{4m+2})=\prod_{n\geq 1}(1-x^{4n})\prod_{m\geq 0}(1+x^{8m+4}) $$ since $(1-x^{2n+1})(1+x^{2n+1})=(1-x^{4n+2})=(1+x^{4n+2})$ and so on. By induction $$ \prod_{n\geq 1}(1-x^n)\prod_{m\geq 0}(1+x^{2m+1})=\prod_{n\geq 1}(1-x^{2^k n})\prod_{m\geq 0}(1+x^{2^{k+1}m+2^k})$$ for any $k\in\mathbb{N}$. It follows that in $\mathbb{F}_2[[x]]$ both series equal $1$, so $p(n)$ and $p_1(n)$ always have the same parity.
By playing a bit with the Jacobi triple product we also have $$ \prod_{n\geq 1}(1-x^n)\prod_{m\geq 0}(1+x^{2m+1})=1+2\sum_{d\geq 1}(-1)^d x^{2d^2}.$$