Why $f(x) = \sqrt{x}$ is a function?

I will assume that in this portion of your textbook it is assumed that $x \in \mathbb{R}$, and with that condition $f(x)=\sqrt{x}$ is certainly a function. Specifically $f:[0,\infty) \rightarrow [0,\infty)$. It meets the formal definition of a function (not one to many).

Your confusion is due to an inappropriate extrapolation of reasoning. Specifically this:

You know that $x^2=25 \Rightarrow x= \pm \sqrt{25}$ by the square root property. However this function in no way involves taking the square root of both sides of an equality. It is just a function $f(x)=\sqrt{x}$, and the domain is $x \geq 0$ by virtue of the fact that you are living in the real number system in this portion of your textbook.

Now $f(x)= \pm \sqrt{x}$ is certainly not a function. For example, if the question were "Let $y^2=x$. Is $y$ a function of $x$?" You would say no in this case as $y= \pm \sqrt{x}$, and you would choose $x=25$ to counter the definition of a function.

Note, your statement "the square root of $25$ has two different outputs" is false. There is only one output. However, if $y^2=25$, then $y$ has two different solutions.

For functions defined by equations, we agree on the following convention regarding the domain: Unless otherwise indicated, the domain is assumed to be the set of all real numbers that lead to unique real-number outputs. The symbol $\displaystyle \sqrt{ }$ is defined in algebra to mean the positive square root only. Thus $\displaystyle \sqrt{25} = 5$, and if you are thinking of the other root, you need to write $\displaystyle -\sqrt{25} = -5$. Consequently, the function $\displaystyle f(x) = \sqrt{x}$ do represent a function; for each input there is exactly one output.