Every continuous map $f:\mathbb{C}P(2) \to \mathbb{C}P(2)$ has a fixed point, without Lefschetz theorem.

Through any two distinct points in $\mathbb{C}P^n$ there is a unique (complex) geodesic. Therefore from any fix point free self map $f$ , we have a 1 dimensional complex subbundle of the tangent bundle by taking the subspace above a point $p$ to be the tangent space of the geodesic from $p$ to $f(p)$ at $p$.

This implies that the total Chern class of $\mathbb{C}P^n$ has a linear factor. If $n=2$ this implies that the total Chern class $1+3x+3x^2$ has two real (integer) roots. However, this is easily checked to be false since the discriminant is negative.

As a sanity check, this should be different if $n=3$. In that case the total Chern class is $1+4x +6x^2 +4x^3$ and this equals $(2 x + 1) (2 x^2 + 2 x + 1)$, as expected.

I imagine if you are better with polynomials than I, you can get this to work for any even $n$.

I don't understand the first paragraph of Connor's answer at all, but assuming it checks out, in general the total Chern class of $\mathbb{CP}^n$ is

$$(1 + \alpha)^{n+1} = \sum_{k=0}^n {n+1 \choose k} \alpha^k \in \mathbb{Z}[\alpha]/\alpha^{n+1}$$

(note that the $k = n+1$ term vanishes). If the (complex) tangent bundle of $\mathbb{CP}^n$ has a (complex) line subbundle then the total Chern class must factor as

$$(1 + c \alpha)(1 + c_1 \alpha + \dots + c_{n-1} \alpha^{n-1})$$

and since we never get a coefficient of $\alpha^{n+1}$ or higher, the problem of determining whether this is possible is equivalent to the problem of determining when $(1 + \alpha)^{n+1} - \alpha^{n+1}$ has a linear factor of the form $(1 + c \alpha)$, as an ordinary polynomial.

It will be easier to reverse the order of the coefficients: this is equivalent to determining when $\frac{(x + 1)^{n+1} - 1}{x}$ has a linear factor of the form $(x + c)$, where $c$ is an integer. This implies

$$(1-c)^{n+1} = 1$$

and if $n$ is even this gives $c = 0$, but $x + c = x$ is not a factor of the above polynomial because its constant coefficient is $n+1$ (this coefficient corresponds to the top Chern class and hence to the Euler characteristic $\chi(\mathbb{CP}^n) = n+1$ so we really are using that the Euler characteristic doesn't vanish). We conclude:

Claim: $\mathbb{CP}^{2m}$ has the fixed point property.

This is usually proven by Lefschetz (as far as I know anyway). If $n$ is odd then this gives either $c = 0$ or $c = 2$ so we don't get a contradiction, and I guess the existence of fixed-point-free maps in this case implies that the tangent bundle has a line subbundle with first Chern class $2$. I wonder if anyone knows a more explicit description of it.