Do I gain any information from removing the random intercept model? I am fitting a mixed effect model to data from a behavioural task where each participant performed the task multiple times, trying to predict a binomial response, something like the formula below:
success ~ var1 + var2 + var 3 + var 4 + (1|participant)
The thing is, I am interested both in the effect of the variables (which can change trial-to-trial) and the individual tendency of each person who did the task.
My very basic understanding leads me to believe that if I were to remove the random intercept term and compare the performance of the one-level model (success ~ var1 + var2 + var 3 + var 4) with the mixed model it would give me some sense of whether the effect of the individual matters to any extent. But I suspect that I might be completely wrong on this.
Am I wrong? And if so, is there any good way to answer my question: is there a considerable effect of individual variability on top of what can be predicted from the 'situational' variables?
 A: 
My very basic understanding leads me to believe that if I were to remove the random intercept term and compare the performance of the one-level model (success ~ var1 + var2 + var 3 + var 4) with the mixed model it would give me some sense of whether the effect of the individual matters to any extent.

This is incorrect.
With your mixed model, the software will estimate a variance for the random intercepts. This will provide inference as to the extent of the overall variation at the participant level.
You will also be able to extract the individual intercepts for each participant and inspect these to help answer your research question. You can
You could also fit a model without random intercepts, but instead fit partitipant as a fixed effect:
success ~ var1 + var2 + var3 + var4 + particpant

and this will give you fixed estimates for the "effect" of each participans along with those for the other fixed effects.
A: To add a bit more context to Robert's post, there is a way to formally test whether the model with the random intercept fits the data better than an OLS model without a random intercept (and no person "fixed effects"). Below is code to demonstrate the idea using the sleepstudy data in the lme4 package.
library(lme4)
data(sleepstudy) 

# First the random intercept model
mlm1 <- lmer(Reaction ~ 1 + (1|Subject), sleepstudy)

# Next the OLS model. 
ols1 <- lm(Reaction ~ 1, data=sleepstudy) 


# Use the anova() function to compute the likelihood ratio test. 
# anova re-fits the OLS model using maximum likelihood

anova(mlm1, ols1) 

refitting model(s) with ML (instead of REML)
Data: sleepstudy
Models:
ols1: Reaction ~ 1
mlm1: Reaction ~ 1 + (1 | Subject)
     npar    AIC    BIC  logLik deviance  Chisq Df Pr(>Chisq)    
ols1    2 1965.0 1971.4 -980.52   1961.0                         
mlm1    3 1916.5 1926.1 -955.27   1910.5 50.508  1  1.187e-12 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

The null hypothesis of this likelihood ratio test is that the random intercept variance is 0; practically this would mean that there is no between person variation in average Reaction time values. The LR test is highly significant, allowing us to reject the null hypothesis. Thus the more complicated lmer model, which differs from the OLS model by 1 parameter (the random intercept variance) provides a better fit to the data. AIC and BIC are both reduced in the lmer model, further supporting the conclusion that the random intercept model does a better job representing the pattern of correlation in the data than the naive OLS model.
