Idea: 0. Every fibre $s^{-1} (x)$ is a closed embedded submanifold of $G$ (as $s$ is a submersion. Thus it makes sense to consider $t|_{s^{-1}(x)}$ (restriction of $t$ to the source fibre) as a smooth map.

1. Prove that $t|_{s^{-1}(x)}$ is a map of constant rank for each $x \in G$.Here Mackenzie uses the existence of enough local bisections (see below)
2. the isotropy group $G_x^x = (t|_{s^{-1}(x)})^{-1}$ is a closed embedded submanifold by the constant rank theorem (see your favorite book on differential geometry or the very general version in https://arxiv.org/pdf/1502.05795.pdf Theorem F). Note that this shows that $G_x^x$ is a submanifold of $s^{-1}(x)$ but since this is an embedded submanifold it is also a submanifold of $G$.
3. This submanifold structure turns $G^x_x$ into a Lie group. Multiplication and inversion are restrictions of the groupoid operations which are smooth on $G$. Now $G_x^x$ is a closed embedded submanifold of $G$ and thus inherits smooth multiplication and inversion from the groupoid.

Now to carry out 1. One constructs first a local bisection through each arrow, i.e. a smooth local section $\sigma \colon U \rightarrow G (U \subseteq M$ open) of $s$ such that $t\circ \sigma$ is a diffeomorphism and $\sigma(s(g))=g$. This exists by some linear algebra and an application of the inverse function theorem. (Details: You take a section of the source map at the point, use linear algebra to show that one can arrange the image of the derivative of the section to be a simultaneous complement to the kernels of the derivatives of source and target map. Then the inverse function theorem yields the desired properties)

Idea:

0. Every fibre $s^{-1} (x)$ is a closed embedded submanifold of $G$ (as $s$ is a submersion). Thus it makes sense to consider $t|_{s^{-1}(x)}$ (restriction of $t$ to the source fibre) as a smooth map.
1. Prove that $t|_{s^{-1}(x)}$ is a map of constant rank for each $x \in G$. Here Mackenzie uses the existence of enough local bisections (see below).
2. the isotropy group $G_x^x = (t|_{s^{-1}(x)})^{-1}(x)$ is a closed embedded submanifold by the constant rank theorem (see your favorite book on differential geometry or the very general version in https://arxiv.org/pdf/1502.05795.pdf Theorem F). Note that this shows that $G_x^x$ is a submanifold of $s^{-1}(x)$ but since this is an embedded submanifold it is also a submanifold of $G$.
3. This submanifold structure turns $G^x_x$ into a Lie group. Multiplication and inversion are restrictions of the groupoid operations which are smooth on $G$. Now $G_x^x$ is a closed embedded submanifold of $G$ and thus inherits smooth multiplication and inversion from the groupoid.

Now to carry out 2. One constructs first a local bisection through each arrow, i.e. a smooth local section $\sigma \colon U \rightarrow G$ ($U \subseteq M$ open) of $s$ such that $t\circ \sigma$ is a diffeomorphism and $\sigma(s(g))=g$. This exists by some linear algebra and an application of the inverse function theorem. (Details: You take a section of the source map at the point, use linear algebra to show that one can arrange the image of the derivative of the section to be a simultaneous complement to the kernels of the derivatives of source and target map. Then the inverse function theorem yields the desired properties).

There is also a rundown in the more modern book of Mackenzie: General Theory of Lie groupoids and Lie algebroids (p.26 Corollary 1.4.11)

Let me summarise the idea: Fix a Lie groupoid $G \rightrightarrows M$ with source map $s$ and target map $t$.

Idea: 1. Prove that $t|_{s^{-1}(x)}$ (restriction of $t$ to the source fibre) is a map of constant rank for each $x \in G$. 2. the isotropy group $G_x^x = (t|_{s^{-1}(x)})^{-1}$ is a closed embedded submanifold by the constant rank theorem. 3. This submanifold structure turns $G^x_x$ into a Lie group (as it inherits smooth multiplication and inversion from the groupoid).

Now to carry out 1. One constructs first a local bisection through each arrow, i.e. a smooth local section $\sigma \colon U \rightarrow G (U \subseteq M$ open) of $s$ such that $t\circ \sigma$ is a diffeomorphism and $\sigma(s(g))=g$. This exists by some linear algebra and an inverse function theorem.

Now this was the finite-dimensional case (in which you are interested I assume). Since I enjoy infinite-dimensional groupoids, let me hijack your question to push this further:

The above proof works as presented for any infinite-dimensional groupoid (using Bastiani calculus to make sense of differentiability beyond Banach spaces) over a finite-dimensional base $M$. See e.g. Appendix A of https://arxiv.org/pdf/1506.05415.pdf for a proof. However, due to the local bisection trick and the constant rank theorem this is as far as one gets.

The nice thing about the Moerdijk, Mrcun approach referenced above is that it can be pushed further. Namely if we are dealing with Banach groupoids (not necessarily finite-dimensional base but now everything is modeled on a Banach space), then this generalised approach yields the Lie group structure of the isotropy groups. This was just recently observed in https://arxiv.org/pdf/1802.09430.pdf (Theorem 3.3 ii)

As far as I know, the question is still open if the isotropy groups of an infinite-dimensional Lie groupoid always inherit a canonical Lie group structure.

There is also a rundown in the more modern book of Mackenzie: General Theory of Lie groupoids and Lie algebroids (p.26 Corollary 1.4.11)

Let me summarise the idea: Fix a Lie groupoid $G \rightrightarrows M$ with source map $s$ and target map $t$.

Idea: 0. Every fibre $s^{-1} (x)$ is a closed embedded submanifold of $G$ (as $s$ is a submersion. Thus it makes sense to consider $t|_{s^{-1}(x)}$ (restriction of $t$ to the source fibre) as a smooth map.

1. Prove that $t|_{s^{-1}(x)}$ is a map of constant rank for each $x \in G$.Here Mackenzie uses the existence of enough local bisections (see below)
2. the isotropy group $G_x^x = (t|_{s^{-1}(x)})^{-1}$ is a closed embedded submanifold by the constant rank theorem (see your favorite book on differential geometry or the very general version in https://arxiv.org/pdf/1502.05795.pdf Theorem F). Note that this shows that $G_x^x$ is a submanifold of $s^{-1}(x)$ but since this is an embedded submanifold it is also a submanifold of $G$.
3. This submanifold structure turns $G^x_x$ into a Lie group. Multiplication and inversion are restrictions of the groupoid operations which are smooth on $G$. Now $G_x^x$ is a closed embedded submanifold of $G$ and thus inherits smooth multiplication and inversion from the groupoid.

Now to carry out 1. One constructs first a local bisection through each arrow, i.e. a smooth local section $\sigma \colon U \rightarrow G (U \subseteq M$ open) of $s$ such that $t\circ \sigma$ is a diffeomorphism and $\sigma(s(g))=g$. This exists by some linear algebra and an application of the inverse function theorem. (Details: You take a section of the source map at the point, use linear algebra to show that one can arrange the image of the derivative of the section to be a simultaneous complement to the kernels of the derivatives of source and target map. Then the inverse function theorem yields the desired properties)

Now this was the finite-dimensional case (in which you are interested I assume). Since I enjoy infinite-dimensional groupoids, let me hijack your question to push this further:

The above proof works as presented for any infinite-dimensional groupoid (using Bastiani calculus to make sense of differentiability beyond Banach spaces) over a finite-dimensional base $M$. See e.g. Appendix A of https://arxiv.org/pdf/1506.05415.pdf for a proof. However, due to the local bisection trick and the constant rank theorem this is as far as one gets.

The nice thing about the Moerdijk, Mrcun approach referenced above is that it can be pushed further. Namely if we are dealing with Banach groupoids (not necessarily finite-dimensional base but now everything is modeled on a Banach space), then this generalised approach yields the Lie group structure of the isotropy groups. This was just recently observed in https://arxiv.org/pdf/1802.09430.pdf (Theorem 3.3 ii)

As far as I know, the question is still open if the isotropy groups of an infinite-dimensional Lie groupoid always inherit a canonical Lie group structure.