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 ${2n \choose n} < 4^n / \sqrt{\pi n}$.

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 $$ \frac{r_n}{r_{n-1}} = \left( 1 - \frac1{2n} \right)^2 \left/ \left( 1 - \frac1n \right) \right. = \frac{(2n-1)^2}{(2n-2)(2n)} \gt 1. $$ 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 $$ R_n := \frac{2n}{2n-1} r_n = \frac{n}{\left(1-\frac 1{2n}\right)16^n} {2n \choose n}^2. $$ 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 ${2n \choose n} < 4^n / \sqrt{\pi n}$.

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 $$ r_n := \left( {2n \choose n} \left/ \frac{4^n}{\sqrt n} \right. \right) = \frac{n}{16^n} {2n \choose n}^2. $$ 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 ${2n \choose n} < 4^n / \sqrt{\pi n}$.

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 ${2n \choose n} < 4^n / \sqrt{\pi n}$.