At least one member of a pythagorean triple is even

Suppose that $a$ and $b$ are both odd, and proceed as you did to conclude that $c$ is even. As $a$ is odd, we may write $a = 2k + 1$ for some $k$; similarly, $b = 2m + 1$. As $c$ is even, write $c = 2n$. This leads to

$$(2k + 1)^2 + (2m + 1)^2 = (2n)^2$$

or upon expanding and regrouping,

$$4(k^2 + k + m^2 + m) + 2 = 4n^2$$

Now the right side is divisible by 4, as is the first term on the left - but $2$ is not divisible by $4$. Do you now see how to derive a contradiction?

Let us try to write this using congruences, which is a very useful and compact notation.

We know that a square can only have remainder $0$ or $1$ modulo $4$. I.e., for any integer $x$ one of these two congruences must be true: $x^2\equiv 0 \pmod 4$ or $x^2\equiv1 \pmod4$. The first case happens if $x$ is even, the second case if $x$ is odd. (Since $(2k+1)^2=4(k^2+k)+1$ and $(2k)^2=4k^2$.)

So we have two possibilities for the remainder of $a^2$ and two possibilities for the remainder of $b^2$ modulo $4$. $$ a^2 \equiv 0 \pmod4, b^2 \equiv 0 \pmod4 \Rightarrow c^2\equiv 0+0=0 \pmod4\\ a^2 \equiv 1 \pmod4, b^2 \equiv 0 \pmod4 \Rightarrow c^2\equiv 1+0=1 \pmod4\\ a^2 \equiv 0 \pmod4, b^2 \equiv 1 \pmod4 \Rightarrow c^2\equiv 0+1=0 \pmod4\\ a^2 \equiv 1 \pmod4, b^2 \equiv 1 \pmod4 \Rightarrow c^2\equiv 1+1=2 \pmod4 $$ We see that in the last case we would have $c^2\equiv 2\pmod4$, which is not possible.

So only the first three cases can really occur. In the other words, either all of the numbers $a$, $b$, $c$ are even, or exactly one of the is even and the remaining two numbers are odd.