Any set with more elements than the dimension of vector space is linearly dependent

There are a constructive and a non constructive proof. The constructive proof is known as “Steinitz Lemma”:

If $B$ is a finite spanning set (in particular, a basis) of a finitely generated vector space $V$ and $A$ is a (finite) linearly independent subset of $V$, then $|A|\le|B|$.

This is proved by showing that there exists a subset $C$ of $B$ with $|C|=|A|$ such that $(B\setminus C)\cup A$ is a spanning set of $V$.

The non constructive proof is maybe easier. Since $\{v_1,\dots,v_n\}$ is a basis of $V$, we can write $$ a_i=\sum_{j=1}^n \gamma_{ij}v_j \qquad(i=1,\dots,p) $$ If $\alpha_1a_1+\dots+\alpha_pa_p=0$, then $$ 0=\sum_{i=1}^p\alpha_ia_i= \sum_{i=1}^p\alpha_i\biggl(\,\sum_{j=1}^n\gamma_{ij}v_j\biggr)= \sum_{j=1}^n\biggl(\,\sum_{i=1}^p\alpha_i\gamma_{ij}\biggr)v_j $$ so $$ \sum_{i=1}^p\alpha_i\gamma_{ij}=0 \qquad(j=1,\dots,n) $$ This is a homogeneous linear system in the unknowns $\alpha_1,\dots,\alpha_p$ and, if $p>n$, it has infinitely many solutions. Therefore $\{a_1,\dots,a_p\}$ is not linearly independent.

There are other proofs, that however depend on knowing more about dimension of (finitely generated) vector spaces.

First fact:

Any (finite) linearly independent set in a vector space $V$ can be extended to a basis.

Second fact:

Any two bases of a vector space have the same number of elements.

Putting together these two facts, you get that a linearly independent set cannot have more elements than a basis.

Hint: If $A$ is linearly independent, then $\dim(\text{span}(A)) = p$, but $\text{span}(A) \subseteq V$. Can you finish the argument?