Your proof provides an interesting use of Hanson's lcm inequality result and appears to be basically correct, but there are few small mistakes & other issues.
In your step (5), you wrote
... $v_p({{3x}\choose{2x}}) = \sum\limits_{i \le \log_p(3n)} \left\lfloor\frac{3x}{p^i}\right\rfloor - \left\lfloor\frac{2x}{p^i}\right\rfloor - \left\lfloor\frac{x}{p^i}\right\rfloor$ which equals $1$ or $0$ for each $i$ ...
First, in the summation part of $i \le \log_p(3n)$, the $n$ I believe is supposed to be $x$. Also, in something like a math journal, the statement "equals $1$ or $0$ for each $i$" may be obvious & sufficient, but I don't believe it is for here since, I at least, didn't initially know if it was always true. FYI, here is what I did to confirm this. I let $d = p^i$ and $x = md + r$ for $m,d \in \mathbb{N}^0$ and $0 \le r \lt d$. Then,
\begin{align}
\left\lfloor \frac{3x}{d} \right\rfloor - \left\lfloor \frac{2x}{d} \right\rfloor - \left\lfloor \frac{x}{d} \right\rfloor & = 3m + \left\lfloor \frac{3r}{d} \right\rfloor - 2m - \left\lfloor \frac{2r}{d} \right\rfloor - m - \left\lfloor \frac{r}{d} \right\rfloor \\
& = \left\lfloor \frac{3r}{d} \right\rfloor - \left\lfloor \frac{2r}{d} \right\rfloor - \left\lfloor \frac{r}{d} \right\rfloor \\
& = \left\lfloor \frac{3r}{d} \right\rfloor - \left\lfloor \frac{2r}{d} \right\rfloor\tag{1}\label{eq1}
\end{align}
Since $r$ and $d$ are positive, both terms are non-negative and $\left\lfloor \frac{3r}{d} \right\rfloor \ge \left\lfloor \frac{2r}{d} \right\rfloor$, so \eqref{eq1} is non-negative. Since $\left\lfloor \frac{3r}{d} \right\rfloor \le 2$, the only way \eqref{eq1} can be anything other than $0$ or $1$ would be for $\left\lfloor \frac{3r}{d} \right\rfloor = 2$ (which requires $\frac{2d}{3} \le r \lt 1$) and $\left\lfloor \frac{2r}{d} \right\rfloor = 0$ (which requires $0 \le r \lt \frac{d}{2}$). Since there is no overlap between those $2$ regions, the result of \eqref{eq1} must always be $0$ or $1$.
Next, you state
If a prime $p > \sqrt{3x}$, then $v_p({{3x}\choose{2x}}) = 1$ and $p | \text{lcm}(\frac{3x}{2})$
For $x = 5$, note that $p = 5 \gt \sqrt{15}$, but ${{3x}\choose{2x}} = {{15}\choose{10}} = 3 \times 7 \times 11 \times 13$, so $v_p({{3x}\choose{2x}}) = 0$ in this case. A correct statement to make would be something like $v_p({{3x}\choose{2x}}) \leq 1$. Note this doesn't really affect your argument and it actually strengthens it slightly as you want to have a sufficiently small upper bound.
With the next statement of
If a prime $p \le \sqrt{3x}$, then $v_p({{3x}\choose{2x}}) \le \log_p(3x) \le v_p(\frac{3x}{2})+1$
first note I believe you mean the last part to be $v_p\left(\text{lcm}\left(\frac{3x}{2}\right)\right) + 1$ instead. However, for $p = 2$ and $x = 4$, then $\log_p(3x) = \log_2(12) \gt 3$, but $v_p\left(\text{lcm}\left(\frac{3x}{2}\right)\right) + 1 = v_2(60) + 1 = 2 + 1 = 3$. A correct statement could use the fact that $v_p$ is always an integer, so you can just use the integer component of the log to get $v_p({{3x}\choose{2x}}) \le \left\lfloor \log_p(3x) \right\rfloor \le v_p\left(\text{lcm}\left(\frac{3x}{2}\right)\right) + 1$.
Finally, for the second part of (7), you wrote
and for $x \ge 246$, $6 > (3^{3/2})(3^{\sqrt{3x+2} - \sqrt{3x}})\left(\dfrac{x+1}{x}\right)$
I don't see how you get this. Your step (6) states that $\dfrac{6^x}{x} < 3^{3x/2}3^{\sqrt{3x}}$. If you multiply by $x$ and take the $x$'th root of both sides, you get $6 \lt 3^{3/2}3^{\sqrt{3/x}}x^{1/x}$. I suggest not only showing how you got your result, but also how you proved it to be always be true for $x \ge 246$.
Here is how I would prove that (6) is not true for $x \ge 246$. First, note that since logarithms, including the natural logarithm, are a strictly increasing function for positive numbers, so $a \ge b \iff \log a \ge \log b$. Thus, take the natural logarithm of both side of (6) and move the right side to the left side to get the following function
\begin{align}
f(x) & = \ln(6)x - \ln(x) - \frac{3\ln(3)}{2}x - \ln(3)\sqrt{3}\sqrt{x} \\
& = \left(\ln(6) - \frac{3\ln(3)}{2}\right)x - \ln(3)\sqrt{3}\sqrt{x} - \ln(x) \tag{2}\label{eq2}
\end{align}
You've already shown that $f(246) \gt 0$. If can show that $f'(x) \ge 0$ for $x \ge 246$, then $f(x) \gt 0$ for all $x \ge 246$. Differentiating \eqref{eq2} gives
\begin{align}
f'(x) & = \left(\ln(6) - \frac{3\ln(3)}{2}\right) - \left(\frac{\ln(3)\sqrt{3}}{2}\right) \frac{1}{\sqrt{x}} - \frac{1}{x} \\
& = \frac{1}{x}\left(\left(\ln(6) - \frac{3\ln(3)}{2}\right)x - \left(\frac{\ln(3)\sqrt{3}}{2}\right)\sqrt{x} - 1\right) \tag{3}\label{eq3}
\end{align}
Note that
$$g(x) = \left(\ln(6) - \frac{3\ln(3)}{2}\right)x - \left(\frac{\ln(3)\sqrt{3}}{2}\right)\sqrt{x} - 1 \tag{4}\label{eq4}$$
is a quadratic equation in $\sqrt{x}$. As the coefficient of $x$, i.e., $\ln(6) - \frac{3\ln(3)}{2} = 0.143841\ldots \gt 0$, the value will always be positive for large enough $x$. Let $y = \sqrt{x}$ to transform \eqref{eq4} to
$$h(y) = \left(\ln(6) - \frac{3\ln(3)}{2}\right)y^2 - \left(\frac{\ln(3)\sqrt{3}}{2}\right)y - 1 \tag{5}\label{eq5}$$
Note that $h(\sqrt{246}) = 19.462358\ldots$. Also,
$$h'(y) = 2\left(\ln(6) - \frac{3\ln(3)}{2}\right)y - \left(\frac{\ln(3)\sqrt{3}}{2}\right) \tag{6}\label{eq6}$$
Since $h'(\sqrt{246}) = 3.560690\ldots$ and $h'(y)$ is a strictly increasing linear function, this shows that, in summary, $f(x)$ in \eqref{eq2} is always positive for $x \ge 246$, which confirms what you're trying to prove.
The answer from mathlove gives several equality counter-examples to your conjecture. I prove below that changing your conjecture's $\lt$ to $\le$ is sufficient to make it correct.
As for your proof, the left part of your expression in step $(3)$ is
$$v_p\left(\frac{(x+n)!}{(x!)\operatorname{lcm}(x+1,\dots,x+n)}\right) = \sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^i}-1\right\rfloor \tag{1}\label{eq1A}$$
With $i$ large enough so $p^i \gt n$, then $-1 \lt \frac{n}{p_i} - 1 \lt 0 \implies \left\lfloor\frac{n}{p^i}-1\right\rfloor = -1$. Thus, $\sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^i}-1\right\rfloor = -\infty$. Also, I believe your note of "It is based on $n$ instead of $x+n$ because $p^t$ is necessarily less than $n$" means you tried using
$$\left\lfloor\frac{x+n}{p^{i}}\right\rfloor - \left\lfloor\frac{x}{p^{i}}\right\rfloor = \left\lfloor\frac{n}{p^{i}}\right\rfloor \tag{2}\label{eq2A}$$
However, even with your restriction, this is not always true. For example, with $t = i = 1$, $p = 3$, $n = 4$ and $x = 5$, the left side of \eqref{eq2A} becomes $\left\lfloor\frac{9}{3}\right\rfloor - \left\lfloor\frac{5}{3}\right\rfloor = 3 - 1 = 2$ while the right side becomes $\left\lfloor\frac{4}{3}\right\rfloor = 1$.
In addition, I assume the $-1$ term on the right side in \eqref{eq1A} comes from the $\operatorname{lcm}$ having its exponent of $p$ be the maximum among the prime factorizations of the integers in $[x+1,x+n]$. However, the $-1$ term should only be used up to that exponent, not to $\infty$, since this causes the sum in \eqref{eq1A} to go to $-\infty$, as I explained earlier.
I don't see any direct way to use what you're trying to do. Instead, your conjecture's $\leq$ version can be proven using an equivalent form of your inequality,
$$\begin{equation}\begin{aligned}
\frac{(x + n)!}{(x!)\operatorname{lcm}(x + 1, \, \ldots, \, x + n)} & \le (n-1)! \iff \\
\frac{(x + n)!}{(x!)(n-1)!} & \le \operatorname{lcm}(x + 1, \, \ldots, \, x + n) \iff \\
n\left(\frac{(x + n)!}{(x!)(n!)}\right) & \le \operatorname{lcm}(x + 1, \, \ldots, \, x + n) \iff \\
n\binom{x + n}{n} & \le \operatorname{lcm}(x + 1, \, \dots, \, x + n)
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Note this form is related to that in your question Reasoning about least common multiples and their relationship to binomial coefficients, and my answer there is also fairly similar to this proof. First, for somewhat simpler algebra, let
$$b = \binom{x + n}{n}, \; c = \operatorname{lcm}(x + 1, \, \dots, \, x + n) \tag{4}\label{eq4A}$$
Using the prime factorization exponents based on the definition of $\operatorname{lcm}$, and the $p$-adic order function, then for any prime $p$, let
$$d = \nu_{p}(n) \tag{5}\label{eq5A}$$
$$e = \nu_{p}(b) \tag{6}\label{eq6A}$$
$$f = \nu_{p}(c) = \max_{x + 1 \, \le \, i \, \le x + n}(\nu_{p}(i)) \tag{7}\label{eq7A}$$
By proving
$$d + e \le f \tag{8}\label{eq8A}$$
for all primes $p$, then \eqref{eq3A} will be true. There are $2$ basic cases to consider:
Case #$1$: $e = 0$
Since $[x + 1, x + n]$ is a set of $n$ consecutive integers, they represent a complete residue system of integers modulo $n$, so there's a $1 \le i \le n$ where $x + i \equiv 0 \pmod n \implies x + i = kn, \; k \in \mathbb{N}$. Thus, for this $i$ and $k$,
$$d = \nu_p(n) \le \nu_p(kn) = \nu_p(x + i) \le f \tag{9}\label{eq9A}$$
which shows \eqref{eq8A} holds.
Case #$2$: $e > 0$
Note Kummer's theorem states $e$ (as defined in \eqref{eq6A}) is the number of carries when $x$ is added to $n$ in each prime base $p$. Thus, there must be at least one carry in this case.
Define for any integer $y \ge 0$ it's standard base $p$ representation as
$$y = \sum_{i=0}^{\infty}a_i p^{i}, \; 0 \le a_i \le p - 1 \tag{10}\label{eq10A}$$
and the following function to specify the requested digit
$$g(y, i) = a_i \tag{11}\label{eq11A}$$
Thus, from \eqref{eq5A},
$$g(n, i) = 0 \; \forall \; 0 \le i \le d - 1 \tag{12}\label{eq12A}$$
i.e., the lowest $d$ digits of $n$ in base $p$ are $0$. This means none of these digits can be involved in a carry when $x$ is added to $n$, so the first possible such digit is $g(n, d)$. Since there's at least one carry, there's a maximum index involved. Let $h \ge d$ be the index of the digit where the highest carry comes from. This means the next higher index digit is increased by $1$, i.e.,
$$\begin{equation}\begin{aligned}
& g(n + x, h + 1) = g(n, h + 1) + g(x, h + 1) + 1 \implies \\
& g(n + x, h + 1) \ge g(x, h + 1) + 1
\end{aligned}\end{equation}\tag{13}\label{eq13A}$$
Note $g(n, h + 1) + g(x, h + 1) + 1 \lt p$ because it can't require a carry since index $h$ is where the highest index of any carry comes from. Also, \eqref{eq13A} shows there's an integer in $[x + 1, x + n]$ where the coefficient at index $h + 1$ increases by $1$ to $g(n + x, h + 1)$, so all lower coefficients then are $0$, i.e., there's a $1 \le i \le n$ where
$$g(n + i, j) = 0 \; \forall \; 0 \le j \le h \implies \nu_p(n + i) \ge h + 1 \tag{14}\label{eq14A}$$
Using \eqref{eq7A}, this means
$$\nu_p(n + i) \le f \implies h + 1 \le f \tag{15}\label{eq15A}$$
The carries occurred only in the digits with indices between $d$ and $h$, inclusive, so the number of carries, i.e., $e$ (as Kummer's theorem mentioned earlier shows), is less or equal to the number of indices in that range. This gives
$$e \le h - d + 1 \implies d + e \le h + 1 \tag{16}\label{eq16A}$$
Putting \eqref{eq15A} and \eqref{eq16A} together gives \eqref{eq8A}.
Best Answer
The calculations are correct.
We recall an inversion formula which can be found for instance as Theorem 268 in An Introduction to the Theory of Numbers by G.H. Hardy and E.M. Wright:
If for all positive $x$ \begin{align*} G(x)=\sum_{j=1}^{\lfloor x\rfloor} F\left(\frac{x}{j}\right)\qquad\text{ then}\qquad F(x)=\sum_{j=1}^{\lfloor x\rfloor}\mu(j)G\left(\frac{x}{j}\right)\tag{3} \end{align*}
Comment:
In (6) we substitute (4) in (5).
In (7) we use the identity $\left\lfloor \left\lfloor \frac{x}{j} \right\rfloor \frac{1}{k}\right\rfloor=\left\lfloor \frac{x}{jk}\right\rfloor$. We also exchange the order of the products which is feasible since they are finite and we set the upper limit to $x$ without changing anything since we are only multiplying with $1$.
In (8) we separate the case $k=1$.
In (9) we divide by $x!$.