Why does the Dirac delta function satisfy $f(x)\delta(x-a) = f(a)\delta(x-a)$?

The definition of the dirac delta is an object $\delta$ such that, for any (suitably well behaved) function $f$, $$\int_{a}^{b}\delta(x)f(x)dx $$ takes the value $f(0)$ if $a<0<b$, and $0$ otherwise. This isn't really a function, as there is provably no function $\delta: \mathbb{R} \to \mathbb{R}$ with this property. If it's not a function, then what the hell is it? Well, as the other answers point out, $\delta$ is formally defined as something called a distribution. However, it turns out you can manipulate $\delta$ as though it were a function, and in surprisingly many cases the sky does not fall on your head.

Fixing an $a \in \mathbb{R}$, it is then immediate from the definition that $$\int_{c}^{d}\delta(x-a)f(x)dx=\int_{c}^{d}f(a)\delta(x-a)dx$$ for any choice of $c$ and $d$, because both sides are equal to $f(a)$ when $a \in [c,d]$, and $0$ otherwise.
It seems that what the question setter has done is then equate the integrands, and say that the objects $$f(x)\delta(x-a)=f(a)\delta(x-a)$$ are the same. In my opinion this is a strange thing to say, because it suggests that you are treating $\delta$ as though it were a real function, which it is not.

Remember that the $\delta$ ''function'' is not really a function (proper or “improper”) but a different mathematical object ( a distribution) that live always in the shade of an integral sign.

So, when we write an ''identity'' as: $$ f(x)\delta(x-a)= f(a)\delta(x-a) , $$ The meaning of this equations is that:

its two sides give equivalent results when used as factors in an integrand.

And this is easy verified using the definition $$ \int_{-\infty}^{+\infty}f(x)\delta(x)dx=f(0)$$ for $x-a=0$.

The only correct way to see it is from the distributional theory of the Delta function. Indeed, for a test function $\phi$, we have:

$$\langle \delta(x-a), \phi(x)\rangle = \langle \theta'(x-a), \phi(x)\rangle = - \langle \theta(x-a), \phi'(x)\rangle = -\int_{-\infty}^{+\infty}\phi'(x)\theta(x-a)\ \text{d}x$$

Now since we have a $\theta$, we know that that $\theta(x-a) = 1$ for $x-a\geq0$ and $0$ otherwise. Hence the integral becomes

$$-\int_{a}^{+\infty}\phi'(x)\theta(x-a)\ \text{d}x = -\phi(x)\bigg|_{a}^{+\infty}$$

The test function $\phi$ goes to zero for $x\to \infty$ hence you remain with

$$\phi(a)$$

This shows you that the action of the $\delta(x-a)$ on a function $f(x)$ gives you

$$f(a)$$

Now the reason why the Dirac Delta remains, it's because the expression $f(x)\delta(x-a)$ makes sense only in the support of the Delta which is $x = a$, otherwise it's zero. To remark this, you write $f(x)\delta(x-a)$ because every other value for $x$ is meaningless.