I believe the difficulty many students have in understanding the proof of
$$
\lim_{x\to0}\frac{\sin x}{x}=1
$$
is that it often comes too early in a calculus course, before one has learned integration, and therefore before one has learned how to think about arc lengths and areas. Since arc length (or area) enters fundamentally into the definition of the $\sin$ function, the proofs that are usually given end up relying on geometric arguments that seem out of place next to the style of argumentation that one otherwise encounters in the first part of a calculus course. I'm not suggesting that these arguments aren't rigorous, or that there aren't good reasons for bringing up trigonometric functions early—but I believe it can be satisfying to understand how this result comes about, rather straightforwardly, once one has defined the trigonometric functions in calculus terms.
If definite integration and arc length are unfamiliar to you, then this proof may not help. However, the notion of arc length is used only to motivate a definition, and all one needs to know about the definite integral is that it represents the area under the curve. This is used to obtain upper and lower bounds.
We make a purely algebraic definition of the $\arcsin$ function as follows. For $y\in(-1,1)$ let
$$\arcsin y=\int_0^y\frac{dt}{\sqrt{1-t^2}}.$$
Interpreted geometrically, this is, in fact, the formula for the signed length of the arc of the unit circle that extends from $(1,0)$ to $(\sqrt{1-y^2},y)$, and therefore indeed the $\arcsin$ function as usually defined, but this definition itself involves no geometry. The $\sin$ function is then the inverse of this function, at least in an interval containing 0. (In fact, in the interval $(-\pi/2,\pi/2),$ where the symbol $\pi$ represents twice the upper bound of the $\arcsin$ function. Thinking about how to extend the function to the entire real line is not needed for our purpose.)
We will first show that
$$
\lim_{y\to0}\frac{\arcsin y}{y}=1.
$$
Since $\sqrt{1-t^2}$ is bounded above by $1,$ the integrand defining $\arcsin y$ is bounded below by $1,$ and, for $y$ positive, $\arcsin y$ is bounded below by $y.$ Hence $(\arcsin y)/y$ is bounded below by $1.$ Since, in the integral, $\sqrt{1-t^2}$ is bounded below by $\sqrt{1-y^2},$ the integrand is bounded above by $1/\sqrt{1-y^2},$ and, for $y$ positive, $\arcsin y$ is bounded above by $y/\sqrt{1-y^2}.$ Hence $(\arcsin y)/y$ is bounded above by $1/\sqrt{1-y^2}.$ Since $(\arcsin y)/y$ is an even function, the same bounds apply for $y$ negative. But both of these bounds have limit $1$ as $y\to0,$ and the result follows from the squeeze theorem.
This is almost the result we want. To finish, define the continuous function
$$
f(y)=\begin{cases}\frac{\arcsin y}{y} & \text{for $y\in(-1,1)\setminus\{0\},$}\\
1 & \text{for $y=0,$}\end{cases}
$$
by plugging the hole at $y=0$ using the limit we just computed. Let $g(x)=\sin x.$ Since $f$ and $g$ are continuous at $0$ and $g(0)=0,$ we have
$$
\lim_{x\to0}f(g(x))=f\left(\lim_{x\to0} g(x)\right)=f(0)=1.
$$
But for $x\ne0,$ we have
$$
f(g(x))=\frac{\arcsin\sin x}{\sin x}=\frac{x}{\sin x},
$$
and so we have our result.
Added remark: Continuity of arcsin on (−1,1) follows in the definition I have used from continuity of the integrand and the Fundamental Theorem of Calculus (which actually implies differentiability). Continuity of sin follows from the easy fact that arcsin is one-to-one on (−1,1) and from the theorem that the inverse of a continuous, one-to-one function is continuous.
