Graph contains a cycle containing at least 4 nodes There is a finite undirected graph in which the degree of every node is at
least 3. 

How to prove that this graph contains a cycle containing at least 4 nodes.

No need to prove that a cycle with a chord, only need to show is that it contains a cycle containing at least 4 nodes.
I am assuming that parallel edges are not allowed.
Firstly, the graph cannot be a tree. Because trees have at least two vertices of degree one (contradict with the assumption). So, as the graph is not a tree, it should contain a cycle.

Let's assume that all cycles in the graph is of length 3. It can be proved that each vertex is at least in one cycle. Using the two previous sentences, the desired graph should be built using cycles of length 3 (triangles). We need to connect triangles to raise the degree of each vertex of triangles. It can be done either by using an edge connecting two vertices of each of the two triangles or by simply attach two vertices from each triangle (one of the vertices should be eliminated). In either case, the two connected vertices would satisfy the required condition (each vertex should be of a degree higher than 3). However, after connecting two triangles, no other edges can be drawn between them, for it makes a cycle of 4. So, raising the degree of the other vertices should be done using more triangles. Now, Let's consider a triangle belonging to the graph. We connect it to three other triangles. The first triangle satisfies the degree condition. However, the other triangles don't. As we cannot connect the edges ,with the degree 2, to the other parts of the current graph (we make a cycle of length more than 3 this way), we need to introduce more triangles and connect them like before. Continuing the process shows that the graph can not be made, for we have vertices of degree two in each iteration. Consequently, the assumption that we made at the beginning of this paragraph, is wrong.
I hope it helps with the main idea.
I am assuming that parallel edges are not allowed.
Firstly, the graph cannot be a tree. Because trees have at least two vertices of degree one (contradict with the assumption). So, as the graph is not a tree, it should contain a cycle.

Let's assume that all cycles in the graph is of length 3. It can be proved that each vertex is at least in one cycle. Using the two previous sentences, the desired graph should be built using cycles of length 3 (triangles). We need to connect triangles to raise the degree of each vertex of triangles. It can be done either by using an edge connecting two vertices of each of the two triangles or by simply attach two vertices from each triangle (one of the vertices should be eliminated). In either case, the two connected vertices would satisfy the required condition (each vertex should be of a degree higher than 3). However, after connecting two triangles, no other edges can be drawn between them, for it makes a cycle of 4. So, raising the degree of the other vertices should be done using more triangles. Now, Let's consider a triangle belonging to the graph. We connect it to three other triangles. The first triangle satisfies the degree condition. However, the other triangles don't. As we cannot connect the edges ,with the degree 2, to the other parts of the current graph (we make a cycle of length more than 3 this way), we need to introduce more triangles and connect them like before. Continuing the process shows that the graph can not be made, for we have vertices of degree two in each iteration. Consequently, the assumption that we made at the beginning of this paragraph, is wrong.
I hope it helps with the main idea.