Continuity of the (real) $\Gamma$ function.

You could also try the basic approach by definition.

For any $\,b>0\,\,\,,\,\,\epsilon>0\,$ choose $\,\delta>0\,$ so that $\,|x-x_0|<\delta\Longrightarrow \left|t^{x-1}-t^{x_0-1}\right|<\epsilon\,$ in $\,[0,b]\,$ : $$\left|\Gamma(x)-\Gamma(x_0)\right|=\left|\lim_{b\to\infty}\int\limits_0^b \left(t^{x-1}-t^{x_0-1}\right)e^{-t}\,dt\right|\leq$$

$$\leq\lim_{b\to\infty}\int\limits_0^b\left|t^{x-1}-t^{x_0-1}\right|e^{-t}\,dt<\epsilon\lim_{b\to\infty}\int\limits_0^b e^{-t}\,dt=\epsilon$$

Since we want to consider continuity at a single fixed point $x_0$, we may assume that $x_0\in(a,b)$ for certain $0<a<1<b$.

Assume $x\in(a,b)$. Using the Mean Value Theorem, there exists $\xi$ between $x$ and $x_0$ such that $$ t^{x-1}-t^{x_0-1}=(\log t)\,t^{\xi-1}\,(x-x_0). $$ So $$ |t^{x-1}-t^{x_0-1}|\leq|(\log t)\,(t^{1-a}+t^{b-1})|\,|x-x_0|. $$ The estimate for $t^{\xi-1}$ is obtained by considering $t<1$ and $t>1$ respectively.

Now \begin{align} \left|\Gamma(x)-\Gamma(x_0)\right| &\leq\int_0^\infty |t^{x-1}-t^{x_0-1}|\,e^{-t}\,dt\\ \ \\ &\leq|x-x_0|\,\int_0^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt\tag{1}\\ \ \\ &\leq\,|x-x_0|\,\left(\frac1{(2-a)^2}+\frac1{b^2}+\frac{2c}{\sqrt e}\right) \end{align}

(see below), and so $\Gamma$ is continuous at $x_0$. Note that $(1)$ is a Lipschitz condition, but it is local: in the sense that the constant depends on $a$ and $b$.

Note that $$\tag{2} \left|\int_0^1 t^r\,\log t\,dt\right|<\infty \iff r>-1. $$

Since $0<a<1$ and $b>0$, we have $1-a>0$ and $b-1>-1$. By $(2)$, then $$\tag{3} \int_0^1|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt\leq-\int_0^1\log t\,(t^{1-a}+t^{b-1})\,dt=\frac1{(2-a)^2}+\frac1{b^2} $$ Let $$c=\sup\{|\log t\,(t^{1-a}+t^{b-1})e^{-t/2}|:\ t\in(0,\infty)\}$$ (the sup exists because the function is continuous and bounded, since both the limits at $0$ and $\infty$ exist). Then \begin{align}\tag{4} \int_1^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt &=\int_1^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t/2}\,e^{-t/2}\,dt\\ \ \\ &\leq c\,\int_1^\infty e^{-t/2}\,dt=\frac{2c}{\sqrt e}. \end{align} Using $(3)$ and $(4)$, we get $$ \int_0^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt\leq\frac1{(2-a)^2}+\frac1{b^2}+\frac{2c}{\sqrt e}. $$

Note that it suffices to show the continuity of $\Gamma(x)$ for all $x>1$ because we can use the known property $\Gamma(x+1)=x\Gamma(x)$ to conclude continuity of those $x$ with $0<x\leq1$. I will break down the proof into two steps:

$1.)$ I show that $\Gamma_k(x):=\int\limits_0^k t^{x-1}e^{-t}dt$ is continuous at point $a>1$.

$2.)$ I show that $\Gamma_k(x)\to\Gamma(x)$ uniformly on a neighbourhood of $a$ so that the limit function inherits the continuity on this neighbourhood.

Proof:

$1.)$ The integrand $K:(1,\infty)\times(0,k)$, where $K(x,t)=t^{x-1}e^{-t}$ is continuous in each point. For a given $\epsilon>0$ and a point $(a,t')$, there exists a $\delta_{t'}$ such that $$\Big\Vert{x\choose t}-{a\choose t'}\Big\Vert<\delta_{t'}\implies | K(x,t)-K(a,t'|<\frac{\epsilon}{2k}.$$ We define a neigbourhood of $t'$ by $$U_{\delta_t'}(t'):=\{t\in(0,k)\mid|t-t'|<\delta_{t'}\}.$$

The union $\bigcup\limits_{t\in(0,k)}U_{\delta_t}(t)$ covers the interval $(0,k)$ and due to compactness there exists a finite cover $\bigcup\limits_{i=1}^nU_{\delta_{t_i}}(t_i)$ of $(0,k)$. Now we set $\delta:=\min\{\delta_1\cdots, \delta_n\}$. Hence, for all $x>1$ with $|x-a|<\delta$ it follows: $$ |\Gamma_k(x)-\Gamma_k(a)|=\left|\int\limits_0^k K(x,t)dt-\int\limits_0^k K(a,t)dt\right|\leq\int\limits_0^k \left|K(x,t)-K(a,t)\right|dt\\ \leq\int\limits_0^k \left|K(x,t)-K(a,t_i)|+|K(a,t_i)-K(a,t)\right|dt<\int\limits_0^k\frac{\epsilon}{k} dt<\epsilon. $$ Here we used that $K$ is continuous in each $(a,t_i)$ and that $t$ lies witihin a $U_{\delta_{t_i}}(t_i)$ so that $\Vert{x\choose t}-{a\choose t_i}\Vert<\delta_i$ and that $\Vert{a\choose t_i}-{a\choose t_i}\Vert<\delta_i$. Hence, $\Gamma_k(x)$ is continuous.

$2.)$ Now let's consider only those $x$ with $1<x<a+1$. Then, we see that $$ \left|\Gamma_k(x)-\Gamma(x)\right|\leq \int\limits_k^{\infty}\left| t^{x-1}e^{-t}dt\right|dt\leq\int\limits_k^{\infty}\left| t^{a}e^{-t}dt\right|dt=\Gamma_k(a+1)-\Gamma(a+1)<\epsilon $$ as $\lim\limits_{k\to\infty}\Gamma_k(a)=\Gamma(a)$. Hence, we have uniform convergence on the neighbourhood of $a$ and $\Gamma(x)$ inherits the continuity at point $a$.