Here's a way to motivate and refine the argument that
Péter Komjáth attributes to Erdős.

Start by computing the ratio between the $n$-th and $(n-1)$-st
central binomial coefficients:
$$
{2n \choose n} \left/ {2(n-1) \choose n-1} \right.
= \frac{(2n)! \phantom. / \phantom. n!^2}{(2n-2)! \phantom. / \phantom. (n-1)^2}
= \frac{(2n)(2n-1)}{n^2}
= 4 \left( 1 - \frac1{2n} \right).
$$
For large $n$, this ratio approaches $4$, suggesting that
$2n \choose n$ grows roughly as $4^n$.
If the factor $1 - \frac1{2n}$ were $1 - \frac1n = (n-1)/n$,
the growth would be exactly proportional to $n^{-1} 4^n$.
Since $1 - \frac1{2n}$ is (for large $n$) nearly the square root of
$1 - \frac1n$, the actual asymptotic should be proportional to $n^{-1/2} 4^n$.
So we introduce the ratio
$$
r_n := \left(
{2n \choose n} \left/ \frac{4^n}{\sqrt n} \right.
\right)^2
= \frac{n}{16^n} {2n \choose n}^2.
$$
Then
$$
\frac{r_n}{r_{n-1}} =
\left( 1 - \frac1{2n} \right)^2
\left/ \left( 1 - \frac1n \right) \right.
= \frac{(2n-1)^2}{(2n-2)(2n)} \lt 1.
$$
Thus $r_{n-1} < r_n$; and since $r_1 = (2/4)^2 = 1/4$ we have by induction
$$
r_1 \lt r_2 \lt r_3 \lt r_4 \lt \cdots \lt r_n
= \frac12
\frac{1 \cdot 3}{2 \cdot 2}
\frac{3 \cdot 5}{4 \cdot 4}
\frac{5 \cdot 7}{6 \cdot 6}
\cdots
\frac{(2n-3)(2n-1)}{(2n-2)(2n-2)}
\frac{2n-1}{2n}.
$$
Each $r_{n_0}$ gives a lower bound on $r_n$, and thus on $2n\choose n$,
for all $n \geq n_0$. The OP asked for upper bounds, so consider
$$
R_n := \frac{2n}{2n-1} r_n
= \frac{n-\frac12}{16^n} {2n \choose n}^2.
$$
Now $R_{n+1}/R_n = (2n-1)(2n+1) \phantom. / \phantom. (2n)^2
= (4n^2-1) \phantom. / \phantom. (4n^2) \lt 1$, so
$$
\frac12 = R_1 \gt R_2 \gt R_3 \gt R_4 \gt \cdots \gt
R_{n+1} =
\frac12
\frac{1 \cdot 3}{2 \cdot 2}
\frac{3 \cdot 5}{4 \cdot 4}
\frac{5 \cdot 7}{6 \cdot 6}
\cdots
\frac{(2n-3)(2n-1)}{(2n-2)(2n-2)}.
$$
It follows that $R_n \geq r_{n'}$ for any $n,n'$, so $R_1=1/2$,
$R_2=3/8$, $R_3=45/128$, etc. are a series of upper bounds on every $r_n$.
Since moreover $r_n / R_n = 1 - \frac1{2n} \rightarrow 1$ as
$n \rightarrow \infty$, both $r_n$ and $R_n$ converge to a common
limit that is an upper bound on every $r_n$. If we accept Wallis's
product (which is classical though not as elementary as everything else
in our analysis), then we can evaluate this common limit as $1/\pi$
and thus recover the asymptotically sharp upper bound
${2n \choose n} < 4^n / \sqrt{\pi n}$.